丁香园AV

Find all critical points and identify them as relative maximum points, relative minimum points, or neither.

1. \(\ds y=x^2-x\)

   Answer
   rel min at \(x=1/2\)
   Solution

   We have \(y' = 2x-1\text{,}\) and so \(y' = 0\) precisely when \(x=1/2\text{.}\) This is the only critical point of \(y\text{.}\) To classify this critical point, we use the First Derivative Test. Since

   \begin{equation*} \begin{split} y'(0) \amp = -1 \lt 0\\ y'(1) \amp = 2-1 = 1 > 0, \end{split} \end{equation*}

   we can construct the following sign diagram:

   

   Therefore, \(f\) must have a relative minimum at \(x=1/2\text{.}\)

2. \(\ds y=2+3x-x^3\)

   Answer
   rel min at \(x=-1\text{,}\) rel max at \(x=1\)
   Solution

   We compute \(y'= 3-3x^2\text{.}\) Hence, \(y'=0\) when \(x=\pm 1\text{.}\) This gives two critical points. Now pick the following test points:

   \begin{equation*} \begin{split} y'(-2) \amp \lt 0 \\ y'(0) \amp > 0\\ y'(2) \amp \lt 0 \end{split} \end{equation*}

   And so we construct the sign diagram:

   

   Therefore, \(y\) has a relative minimum at \(x=-1\) and a relative maximum at \(x=1\text{.}\)

3. \(\ds y=x^3-9x^2+24x\)

   Answer
   rel max at \(x=2\text{,}\) rel min at \(x=4\)
   Solution

   Differentiating, we find \(y'=3x^2-18x+24\text{.}\) Therefore, \(y'=0\) at \(x=2\) and \(x=4\text{.}\) To classify these two critical points, we pick the following test points:

   \begin{equation*} \begin{split} y'(1) \amp > 0\\ y'(3) \amp \lt 0\\ y'(5) \amp >0 \end{split} \end{equation*}

   Now construct the following sign diagram:

   

   Thus, by the First Derivative Test, \(y\) has a relative minimum at \(x=4\) and a relative maximum at \(x=2\text{.}\)

4. \(\ds y=x^4-2x^2+3\)

   Answer
   rel min at \(x=\pm 1\text{,}\) rel max at \(x=0\)
   Solution

   We have \(y'(x)=4x^3-4x = 4x(x^2-1)\text{,}\) and so \(y'=0\) when \(x=-1,0,1\text{.}\) To classify these critical points, we use the first derivative test. We pick the following test points:

   \begin{equation*} \begin{split} y'(-2) \amp = -8(4-1) \lt 0 \\ y'(-0.5) \amp = -2(0.25-1) > 0 \\ y'(0.5) \amp = 2(0.25-1) \lt 0 \\ y'(2) \amp = 8(4-1) > 0 \end{split} \end{equation*}

   And so we can construct the following sign diagram for \(y\text{:}\)

   

   Hence, \(y\) has relative minima at \(x=\pm 1\) and a relative maximum at \(x=0\text{.}\)

5. \(\ds y=3x^4-4x^3\)

   Answer
   rel min at \(x=1\)
   Solution

   Differentiating, we find \(y' = 12x^3-12x^2\text{.}\) Therefore, \(y'=0\) when \(x=0,1\text{.}\) Since

   \begin{equation*} \begin{split} y'(-1) \amp \lt 0 \\ y'(0.5) \amp \lt 0\\ y'(2) \amp >0 \end{split} \end{equation*}

   we can construct the following sign diagram:

   

   Hence, \(y\) has a relative minimum at \(x=1\text{,}\) and neither a relative maximum nor minimum at \(x=0\text{.}\)

6. \(\ds y=(x^2-1)/x\)

   Answer
   none
   Solution

   First notice that the domain of \(y\) is \(\mathbb{R} \setminus \{0\}\text{.}\) Next, we differentiate:

   \begin{equation*} \diff{y}{x} = \diff{}{x} \frac{x^2-1}{x} = \frac{1}{x^2}+1\text{.} \end{equation*}

   Since

   \begin{equation*} y' = \frac{1}{x^2}+1= 0 \end{equation*}

   has no real solutions, and \(y'\) is defined everywhere in the domain of \(y\text{,}\) \(y\) has no critical points.

7. \(\ds y=3x^2-(1/x^2)\)

   Answer
   none
   Solution

   Notice that the domain of \(y\) is \(\mathbb{R} \setminus \{0\}\text{.}\) Next, we differentiate:

   \begin{equation*} y' = 6x + \frac{2}{x^3}\text{.} \end{equation*}

   Then

   \begin{equation*} 6x + \frac{2}{x^3} = 0 \implies x^4 = -\frac{3}{2}\text{,} \end{equation*}

   and so this gives no critical points. Although \(y'\) is undefined at \(x=0\text{,}\) this is not a point in the domain and so \(y\) has no critical points, and hence no relative extrema.

8. \(\ds y=\cos(2x)-x\)

   Answer
   rel min at \(x=7\pi/12+k\pi\text{,}\) rel max at \(x=-\pi/12+k\pi\text{,}\) for integer \(k\)
   Solution

   The domain of \(y\) is all real numbers. Now differentiate:

   \begin{equation*} y' = -2\sin(2x)-1\text{.} \end{equation*}

   Hence, the critical points of \(y\) are given by the solutions to

   \begin{equation*} y'=0 \implies \sin(2x) = \frac{1}{2}\text{.} \end{equation*}

   The solution to

   \begin{equation*} \sin^{-1} \frac{1}{2} = 2x \end{equation*}

   is \(x = \frac{\pi}{12}\text{.}\) This gives the solution in the interval \([0, \pi/2]\text{.}\) Thus, \(x = \frac{\pi}{2} - \frac{\pi}{12} = \frac{5\pi}{12}\) is the second solution over the entire interval \([0,\pi]\text{.}\) Therefore, there are an infinite number of critical points:

   \begin{equation*} x = \frac{\pi}{12} + k\pi \text{ and } x = \frac{5\pi}{12} + k\pi\text{,} \end{equation*}

   for any integer \(k\text{.}\) We classify the critical points over one period of \(\cos 2x\text{,}\) \([0,\pi]\text{.}\) Since

   \begin{equation*} \begin{split} f'(0) \amp \lt 0 \\ f'(3\pi/12) \amp > 0 \\ f'(\pi) \amp \lt 0 \end{split} \end{equation*}

   we conclude that \(y\) has relative minima at \(x=\frac{\pi}{12}+k\pi\) and relative maxima at \(x=\frac{5\pi}{12}+k\pi\text{.}\)

9. \(\ds f(x) = (5-x)/(x+2)\)

   Answer
   none
   Solution

   The domain of \(y\) is all real numbers such that \(x \neq -2\text{.}\) Now differentiate:

   \begin{equation*} y' = \frac{-7}{(x+2)^2}\text{.} \end{equation*}

   The derivative \(y'\) is defined over the entire domain of \(y\text{,}\) so the critical points of \(y\) are given by

   \begin{equation*} y'=0 \implies -7 = 0\text{,} \end{equation*}

   which has no solutions. Therefore, \(y\) has no relative maxima or minima.

10. \(\ds f(x) = |x^2 - 121|\)

    Answer
    rel max at \(x=0\text{,}\) rel min at \(x=\pm 11\)
    Solution

    We first write \(f\) using the piecewise definition of the absolute value function.

    \begin{equation*} f(x) = \begin{cases}121 - x^2 \amp \text{ if } -11 \leq x \leq 11 \\ x^2 - 121 \amp \text{ otherwise } \end{cases} \end{equation*}

    The graph of \(f\) will have two corners: one at \(x=11\text{,}\) and one at \(x=-11\text{.}\) We find

    \begin{equation*} f'(x) = \begin{cases}-2x \amp \text{ if } -11 \lt x \lt 11 \\ 2x \amp \text{ otherwise } \end{cases} \end{equation*}

    where \(f'\) is undefined at \(x= \pm 11\text{.}\) Since \(f'(x) = 0\) at \(x=0\text{,}\) we therefore have 3 critical points, namely \(x=-11, 0\) and \(11\text{.}\) We will classify these critical points using the first derivative test. We pick the following test points and construct a sign diagram for \(f'\text{:}\)

    \begin{equation*} \begin{split} f'(-12) \amp = 2(-12) \lt 0 \\ f'(-10) \amp = -2(-10) > 0 \\ f'(10) \amp = -2(10) \lt 0 \\ f'(12) \amp = 2(12) > 0 \end{split} \end{equation*}

    

    Therefore, \(x=0\) corresponds to a relative maximum, and \(x=\pm 11\) correspond to relative minima of \(f\text{.}\)

11. \(\ds f(x) = x^3/(x+1)\)

    Answer
    rel min at \(x=-3/2\text{,}\) neither at \(x=0\)
    Solution

12. \(\ds f(x) = \sin ^2 x\)

    Answer
    rel min at \(n\pi\text{,}\) rel max at \(\pi/2+n\pi\)
    Solution

    First, we note that the domain of \(f(x)=\dfrac{x^3}{(x+1)}\) is \((-\infty,-1) \cup (-1,\infty)\text{.}\) Now,

    \begin{equation*} f'(x)=\frac{x^2(2x+3)}{(x+1)^2}\text{,} \end{equation*}

    so \(f'\) is also undefined at \(x=-1\text{,}\) but this is not a critical point as it is not in the domain of \(f\text{.}\) To find the critical points of \(f\text{,}\) we solve

    \begin{equation*} \begin{split} f'(x) \amp = 0 \\ x^2(2x+3) \amp = 0 \\ x \amp = 0,-3/2. \end{split} \end{equation*}

    We pick appropriate test points at which to evaluate \(f'\text{,}\) for example

    \begin{equation*} f'(-2)= -4, \ \ f'(-0.5)= 2, \ \ \text{ and } \ \ f'(1)=5/4\text{.} \end{equation*}

    Thus, we find the following sign diagram for \(f'\text{:}\)

    

    Thus, \(f\) has a relative minimum at \(x=-3/2\text{,}\) and the critical point \(x=0\) is neither a relative maximum nor a relative minimum.

13. \(\ds f(x)=\sec x\)

    Answer
    rel min at \(2n\pi\text{,}\) rel max at \((2n+1)\pi\text{,}\) for integer \(n\)
    Solution

    The function \(f(x)=\sin^2(x)\) is defined for all real numbers and has a period of \(\pi\text{.}\) Its derivative,

    \begin{equation*} f'(x)=2\sin x \cos x \end{equation*}

    is also defined for all real numbers and has infinitely many zeroes as follows.

    \begin{equation*} f'(x) = 0 \implies \sin x = 0 \ \ \text{ or } \ \ \cos x = 0 \implies x = \frac{n\pi}{2}\text{,} \end{equation*}

    for all integers \(n\text{.}\) It suffices to classify the critical points of \(f\) over one period, for example \([0,\pi]\text{.}\) We find the sign diagram of \(f'\) to be

    

    We see that \(x=0\) and \(x=\pi\) are relative minima, and \(x=\pi/2\) is a relative maximum. To generalize these results, we split the critical points \(x=\frac{n\pi}{2}\) into two cases: \(n\) odd, and \(n\) even. We can represent \(n\) even as \(n=2m\) for all integers \(m\text{,}\) and \(n\) odd as \(n=2m+1\) for all integers \(m\text{.}\) Thus, \(x=m\pi\) are relative minima and \(x = \pi/2+m\pi\) are relative maxima.

Let \(\ds f(\theta) = \cos^2(\theta) - 2\sin(\theta)\text{.}\) Find the intervals where \(f\) is increasing and the intervals where \(f\) is decreasing in \([0,2\pi]\text{.}\) Use this information to classify the critical points of \(f\) as either relative maximums, relative minimums, or neither.

Let \(r>0\text{.}\) Find the relative maxima and minima of the function \(\ds f(x) =\sqrt{r^2 -x^2 }\) on its domain \([-r,r]\text{.}\)

Given the graph of a function \(f\text{,}\) determine the intervals where \(f\) is increasing, constant, and decreasing.

1. 
   Answer
   decreasing on \((-\infty,0)\text{,}\) increasing on \((0,\infty)\)

2. 
   Answer
   increasing on \((-1,1)\cup(3,\infty)\) and decreasing on \((-\infty,-1)\cup(1,3)\text{.}\)

3. 
   Answer
   increasing on \((-2,-1)\text{,}\) decreasing on \((1,2)\text{,}\) constant on \((-1,1)\)

4. 
   Answer
   decreasing on \((-\infty,1)\cup(1,2)\) and increasing on \((2,\infty)\text{.}\)

Given the graph of a function \(f\text{,}\) determine the relative maxima and relative minima, if any.

1. 
   Answer
   rel min at -1,1 and rel max at 0

2. 
   Answer
   rel min at -1,2 and rel max at 0

3. 
   Answer
   rel min at -1,1 and rel max at 0

4. 
   Answer
   rel min at 4 and rel max at -2

A subsidiary of ThermoMaster manufactures an indoor-outdoor thermometer. Management estimates that the profit (in dollars) realizable by the company for the manufacture and sale of \(x\) units of thermometers each week is

Find the intervals where the profit function \(P\) is increasing and the intervals where \(P\) is decreasing.

Sales in the Web-hosting industry are projected to grow in accordance with the function

where \(f(t)\) is measured in billions of dollars and \(t\) is measured in years, with \(t=0\) corresponding to 1999.

1. Find the interval where \(f\) is increasing and the interval where \(f\) is decreasing. Answer
   Increasing on \((0,6)\)
   Solution

   To find the intervals of increase and decrease, we need to find the zeros of \(f'\text{.}\)

   \begin{equation*} f'(x) = -0.15t^2+1.12t+5.47 = 0\text{,} \end{equation*}

   which gives \(t=-3.36628\) and \(t=10.8329\text{.}\) \(f\) therefore has no critical points since neither of these two solutions are in its domain. We check the sign of \(f'\) by picking any point in \([0,6]\text{,}\) e.g \(f(0) = 5.47 > 0\text{,}\) and so \(f\) is increasing on the entire interval \([0,6]\text{.}\)

2. What does your result tell you about the sales in the Web-hosting industry from 1999 through 2005? Answer

   We conclude that sales were increasing from 1999 to 2005.

1. Find the interval where \(N\) is increasing and the interval where \(N\) is decreasing. Answer
   Increasing on \((0,6)\text{.}\)
   Solution

   We first compute \(f'(t)\text{:}\)

   \begin{equation*} N'(t) = 3(0.09444)t^2-2(1.44167)t+10.65695 = 0.28332t^2-2.8833tt+10.65695 \end{equation*}

   Using the quadratic formula, we find that \(N'(t) = 0\) has no real solutions. Since \(N'(0) = 10.65695 > 0\text{,}\) we see that \(N(t)\) is increasing on the entire interval \([0,6]\text{.}\)

2. What does your result tell you about the number of subscribers in the Canadian cellular market in the years under consideration? Answer

   This means that the number of subscribers is increasing from 1997 to 2003.

Let \(\ds f(x) =a x^2 + bx + c\) with \(a\neq 0\text{.}\) Show that \(f\) has exactly one critical point using the First Derivative Test. Give conditions on \(a\) and \(b\) which guarantee that the critical point will be a maximum.

State the domain and determine the vertical and horizontal asymptotes, if any, of the following functions.

1. \(f(x)=\dfrac{9x-1}{x^2-9}\)

   Answer
   \(D=(-\infty,-3)\cup(-3,3)\cup(3,\infty)\text{;}\)\(x=\pm 3\text{;}\)\(y=0\text{.}\)
   Solution

   The domain of \(f\) is all real numbers such that

   \begin{equation*} x^2-9 \neq 0 \implies x \neq \pm 3\text{.} \end{equation*}

   Therefore, the domain of \(f\) is \((-\infty,-3)\cup(-3,3)\cup(3,\infty)\text{.}\) We now look for vertical asymptotes:

   \begin{equation*} \lim_{x\to-3^-} \frac{9x-1}{x^2-9} = \lim_{x\to-3^-} \frac{9x-1}{(x-3)(x+3)} = -\infty\text{,} \end{equation*}

   and similarly

   \begin{equation*} \lim_{x\to-3^+} \frac{9x-1}{x^2-9} = \lim_{x\to-3^+} \frac{9x-1}{(x-3)(x+3)} = \infty\text{.} \end{equation*}

   Therefore, \(f(x)\) has a vertical asymptote at \(x=-3\text{.}\) We now check the behaviour of the function around \(x=3\text{:}\)

   \begin{equation*} \lim_{x\to 3^-} \frac{9x-1}{x^2-9} = \lim_{x\to 3^-} \frac{9x-1}{(x-3)(x+3)} = -\infty\text{,} \end{equation*}

   and similarly

   \begin{equation*} \lim_{x\to 3^+} \frac{9x-1}{x^2-9} = \lim_{x\to 3^+} \frac{9x-1}{(x-3)(x+3)} = \infty\text{.} \end{equation*}

   So \(f(x)\) has a second vertical asymptote at \(x=3\text{.}\) We now look for horizontal asymptotes. First, check

   \begin{equation*} \lim_{x\to\infty} \frac{9x-1}{x^2-9} = \lim_{x\to\infty} \frac{9x}{x^2} = \lim_{x\to\infty} \frac{9}{x} = 0\text{.} \end{equation*}

   Similarly, we find

   \begin{equation*} \lim_{x\to-\infty} \frac{9x-1}{x^2-9} = \lim_{x\to-\infty} \frac{9x}{x^2} = \lim_{x\to-\infty} \frac{9}{x} = 0\text{.} \end{equation*}

   Therefore, the graph of the function \(f(x)\) has one horizontal asymptote at \(y=0\text{.}\)

2. \(f(x)=\dfrac{8x^2+9}{x^2-16}\)

   Answer
   \(D=(-\infty,-4)\cup(-4,4)\cup(4,\infty)\text{;}\)\(x=\pm 3\text{;}\)\(y=8\text{.}\)
   Solution

   The domain of \(f(x)\) is \((-\infty,-4)\cup(-4,4)\cup(4,\infty)\text{.}\) We first check for vertical asymptotes. Since

   \begin{equation*} \lim_{x\to-4^-} \frac{8x^2+9}{(x-4)(x+4)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to-4^+} \frac{8x^2+9}{(x-4)(x+4)} = \infty\text{,} \end{equation*}

   we conclude that \(f(x)\) has a vertical asymptote at \(x=-4\text{.}\) Similarly,

   \begin{equation*} \lim_{x\to 4^-} \frac{8x^2+9}{(x-4)(x+4)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to 4^+} \frac{8x^2+9}{(x-4)(x+4)} = \infty\text{,} \end{equation*}

   and so \(f(x)\) has a second vertical asymptote at \(x=4\text{.}\) To look for any horizontal asymptotes, we compute:

   \begin{equation*} \lim_{x\to \infty} \frac{8x^2+9}{x^2-16} = \lim_{x\to \infty} \frac{8x^2}{x^2} = 8\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to -\infty} \frac{8x^2+9}{x^2-16} = \lim_{x\to -\infty} \frac{8x^2}{x^2} = 8\text{.} \end{equation*}

   Therefore, the graph of \(f(x)\) has one horizontal asymptote at \(y=8\text{.}\)

3. \(f(x)=\dfrac{6(x^2+10)}{x^2+2x-15}\)

   Answer
   \(D=(-\infty,-5)\cup(-5,3)\cup(3,\infty)\text{;}\)\(x=-5,3\text{;}\)\(y=6\text{.}\)
   Solution

   The domain of \(f(x)\) is \((-\infty,-5)\cup(-5,3)\cup(3,\infty)\text{.}\) First, we will check for any vertical asymptotes:

   \begin{equation*} \lim_{x\to -5^-} \frac{6(x^2+10)}{(x-3)(x+5)} = -\infty \end{equation*}

   and

   \begin{equation*} \lim_{x\to -5^+} \frac{6(x^2+10)}{(x-3)(x+5)} = +\infty\text{.} \end{equation*}

   Similarly, we find that

   \begin{equation*} \lim_{x\to 3^-} \frac{6(x^2+10)}{(x-3)(x+5)} = -\infty \end{equation*}

   and

   \begin{equation*} \lim_{x\to 3^+} \frac{6(x^2+10)}{(x-3)(x+5)} = +\infty\text{.} \end{equation*}

   Therefore, the graph of \(f(x)\) has two vertical asymptotes, one at \(x=3\) and another at \(x=-5\text{.}\) Next, we compute the following limits:

   \begin{equation*} \lim_{x\to \infty} \frac{6(x^2+10)}{x^2+2x-15} = \lim_{x\to\infty} \frac{6x^2}{x^2} = 6\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to -\infty} \frac{6(x^2+10)}{x^2+2x-15} = \lim_{x\to-\infty} \frac{6x^2}{x^2} = 6\text{.} \end{equation*}

   Thus, the graph of \(f(x)\) has one horizontal asymptote at \(y=6\text{.}\)

Match the graphs labelled A through F with one of the rational functions labelled \(a\) through \(f\text{.}\)

Draw and label the vertical and horizontal asymptotes, if any, directly on the graphs.

1. 
   Answer
   HA at \(y=0\)
   Solution

   Since

   \begin{equation*} \lim_{x\to\infty} \left(2-\frac{1}{x^2}\right) = 2\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to 0 } \left(2-\frac{1}{x^2}\right) = - \infty\text{,} \end{equation*}

   \(y\) has a horizontal asymptote at \(y=2\) and a vertical asymptote at \(x=0\text{.}\)

   

2. 
   Answer
   HA at \(y=0\)
   Solution

   Since \(\sqrt{x^2+1} = 0\) has no solutions, \(y\) has no vertical asymptotes. Now, we see that

   \begin{equation*} \lim_{x\to -\infty} \frac{3x}{\sqrt{x^2+1}} = -3, \text{ and }\text{,} \end{equation*}

   \begin{equation*} \lim_{x\to \infty} \frac{3x}{\sqrt{x^2+1}} = 3\text{.} \end{equation*}

   Hence, \(y\) has two horizontal asymptoptes: at \(y=\pm 3\text{.}\)

   

3. 
   Answer
   VA at \(x=\pm 1\text{,}\) HA at \(y=0\)

4. 
   Answer
   VA at \(x=0\text{,}\) HA at \(y=2\)

5. 
   Answer
   HA at \(y=\pm 3\)

6. 
   Answer

Determine the horizontal and vertical asymptotes of the following functions.

1. \(f(x)=\dfrac{1}{x}\)

   Answer
   HA at \(y=0\text{;}\) VA at \(x=0\)
   Solution

   Since

   \begin{equation*} \lim_{x\to 0^-} \frac{1}{x} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to 0^+} \frac{1}{x} = \infty\text{,} \end{equation*}

   we see that \(f(x)\) has a vertical asymptote at \(x=0\text{.}\) Furthermore,

   \begin{equation*} \lim_{x\to\infty} \frac{1}{x} = \lim_{x\to-\infty}\frac{1}{x} = 0\text{,} \end{equation*}

   and so \(f(x)\) has a horizontal asymptote at \(y=0\text{.}\)

2. \(f(q) = -\dfrac{2}{q^2}\)

   Answer
   HA at \(y=0\text{;}\) VA at \(x=0\)
   Solution

   Since

   \begin{equation*} \lim_{q\to 0^-} \frac{-2}{q^2} =\lim_{q\to 0^+} \frac{-2}{q^2} = -\infty\text{,} \end{equation*}

   \(f(q)\) has a vertical asymptote at \(q=0\text{.}\) Next, since

   \begin{equation*} \lim_{q\to\infty} \frac{-2}{q^2} = \lim_{q\to-\infty}\frac{-2}{q^2} = 0\text{,} \end{equation*}

   and so \(f(q)\) has a horizontal asymptote at \(y=0\text{.}\)

3. \(g(t) = \dfrac{t-1}{t+1}\)

   Answer
   HA at \(y=1\text{;}\) VA at \(x=-1\)
   Solution

   We investigate the behaviour of \(g\) as \(t\) appraoched \(-1\text{:}\)

   \begin{equation*} \lim_{t\to-1^+} \frac{t-1}{t+1} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{t\to-1^-} \frac{t-1}{t+1} = \infty\text{.} \end{equation*}

   Hence, \(g(t)\) has a vertical asymptote at \(t=-1\text{.}\) We now compute

   \begin{equation*} \lim_{t\to\infty} \frac{t-1}{t+1} = \lim_{t\to\infty} \frac{t}{t} = 1\text{.} \end{equation*}

   Similarly, we find

   \begin{equation*} \lim_{t\to -\infty} \frac{t-1}{t+1} = \lim_{t\to-\infty} \frac{t}{t} = 1\text{.} \end{equation*}

   Therefore, \(g(t)\) has a horizontal asymptote at \(t=1\text{.}\)

4. \(h(s) = s^3-3s^2+s+1\)

   Answer
   none
   Solution

   Since the function \(h(s)\) is a polynomial, it has no horizontal nor vertical asymptotes.

5. \(f(t) = \dfrac{t^2}{t^2-9}\)

   Answer
   HA at \(y=1\text{;}\) VA at \(t= \pm 3\)
   Solution

   First, we notice that we can rewrite

   \begin{equation*} f(t) = \frac{t^2}{(t-3)(t+3)}\text{.} \end{equation*}

   So we investigate:

   \begin{equation*} \lim_{t\to 3^-} \frac{t^2}{(t-3)(t+3)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{t\to 3^+} \frac{t^2}{(t-3)(t+3)} = \infty\text{.} \end{equation*}

   Similarly,

   \begin{equation*} \lim_{t\to -3^-} \frac{t^2}{(t-3)(t+3)} = \infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{t\to -3^+} \frac{t^2}{(t-3)(t+3)} = -\infty\text{.} \end{equation*}

   We conclude that \(f(t)\) has two vertical asymptotes at \(t=\pm 3\text{.}\) We now look for any horizontal asymptotes:

   \begin{equation*} \lim_{t\to\infty} \frac{t^2}{t^2-9} = \lim_{t\to\infty} \frac{t^2}{t^2} = 1\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{t\to-\infty} \frac{t^2}{t^2-9} = \lim_{t\to-\infty} \frac{t^2}{t^2} = 1\text{,} \end{equation*}

   and so the graph of \(f\) as one horiztonal asymptote at \(y=1\text{.}\)

6. \(f(x) = \dfrac{3x}{x^2-x-6}\)

   Answer
   HA at \(y=0\text{;}\) VA at \(x=-2,3\)
   Solution

   We first factor the denominator:

   \begin{equation*} f(x) = \frac{3x}{(x+2)(x-3)}\text{.} \end{equation*}

   Now investigate the behaviour as \(x\) approches \(-2\text{:}\)

   \begin{equation*} \lim_{x\to-2^-} \frac{3x}{(x+2)(x-3)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to-2^+} \frac{3x}{(x+2)(x-3)} = \infty\text{.} \end{equation*}

   Hence, the graph of \(f(x)\) has a vertical asymptote at \(x=-2\text{.}\) Next, since

   \begin{equation*} \lim_{x\to 3^-} \frac{3x}{(x+2)(x-3)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to 3^+} \frac{3x}{(x+2)(x-3)} = \infty\text{,} \end{equation*}

   we conclude that \(f(x)\) has a second vertical asymptote at \(x=3\text{.}\) We now look for any horizontal asymptotes:

   \begin{equation*} \lim_{x\to\infty} \frac{3x}{x^2-x-6} = \lim_{x\to\infty} \frac{3x}{x^2} = 0\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{x\to-\infty} \frac{3x}{x^2-x-6} = \lim_{x\to-\infty} \frac{3x}{x^2} = 0\text{.} \end{equation*}

   Therefore, the graph of \(f(x)\) has one horizontal asymptote at \(y=0\text{.}\)

7. \(h(t)=2+\dfrac{5}{(t-2)^2}\)

   Answer
   HA at \(y=2\text{;}\) VA at \(t=2\)
   Solution

   We first investigate the behaviour of \(h\) as \(t\) approached \(2\text{.}\)

   \begin{equation*} \lim_{t\to 2} \left(2+\frac{5}{(t-2)^2}\right) = 2+ 5\lim_{t\to 2} \frac{1}{(t-2)^2} = \infty\text{.} \end{equation*}

   Therefore, the graph of \(h(t)\) has a vertical asymptote at \(t=2\text{.}\) Since

   \begin{equation*} \lim_{t\to\infty} \left(2+\frac{5}{(t-2)^2}\right) = 2\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{t\to-\infty} \left(2+\frac{5}{(t-2)^2}\right) = 2\text{,} \end{equation*}

   we see that the graph of \(h(t)\) has one horizontal asymptote at \(y=2\text{.}\)

8. \(f(s)=\dfrac{s^2-2}{s^2-4}\)

   Answer
   HA at \(y=1\text{;}\) VA at \(x= \pm 2\)
   Solution

   First, notice that

   \begin{equation*} f(s) = \frac{s^2-2}{s^2-4} = \frac{s^2-2}{(s+2)(s-2)}\text{.} \end{equation*}

   So as \(s\) approaches \(\pm 2\text{,}\) we find:

   \begin{equation*} \lim_{s\to -2^-} \frac{s^2-2}{(s+2)(s-2)} = \infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{s\to -2^+} \frac{s^2-2}{(s+2)(s-2)} = -\infty\text{,} \end{equation*}

   which means that the graph of \(f\) has a vertical asymptote at \(s=-2\text{.}\) Furthermore, we find that

   \begin{equation*} \lim_{s\to 2^-} \frac{s^2-2}{(s+2)(s-2)} = -\infty\text{,} \end{equation*}

   and

   \begin{equation*} \lim_{s\to 2^+} \frac{s^2-2}{(s+2)(s-2)} = +\infty\text{,} \end{equation*}

   and so the graph of \(f\) has a second vertical asymptote at \(s=2\text{.}\) We now look for any horizontal asymptotes. To do this, we compute

   \begin{equation*} \lim_{s\to\infty} \frac{s^2-2}{s^2-4} = \lim_{s\to\infty} \frac{s^2}{s^2} = 1\text{,} \end{equation*}

   which is additionally equal to

   \begin{equation*} \lim_{s\to-\infty} \frac{s^2-2}{s^2-4} = \lim_{s\to-\infty} \frac{s^2}{s^2} = 1\text{.} \end{equation*}

   We conclude that \(f\) has one horizontal asymptote at \(y=1\text{.}\)

9. \(g(x) = \dfrac{x^3-x}{x(x+1)}\)

   Answer
   none
   Solution

   We first notice that we can factor \(g(x)\) as follows:

   \begin{equation*} g(x) = \frac{x^3-x}{x(x+1)} = \frac{x(x-1)(x+1)}{x(x+1)}\text{.} \end{equation*}

   Therefore, \(g\) has removable (or hole) discontinuities at \(x=0\) and \(x=-1\text{,}\) and hence no vertical asymptotes. Furthermore,

   \begin{equation*} \lim_{x\to \pm \infty} g(x) = \lim_{x\to \pm \infty} (x-1) = \lim_{x \to \pm \infty} x\text{,} \end{equation*}

   and so \(g(x)\) either grows or decreases without bound as \(x \to \pm \infty\text{,}\) respectively. Thus, \(g\) has no horizontal asymptotes.

Find the slant asymptote of the following functions:

1. \(f(x)=\dfrac{5x^2-3x+1}{x+2}\)

   Answer
   \(y=5x+13\)
   Solution

   We wish to find a line \(y=mx+b\) such that

   \begin{equation*} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (mx+b)\right] = 0\text{.} \end{equation*}

   We write:

   \begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{5x^2-3x+1 - mx^2-(2m+b)x-2b}{x+2}\\ \amp = \lim_{x\to \pm \infty} \frac{5x-3+1/x - mx-(2m+b)-2b/x}{1+2/x}. \end{split} \end{equation*}

   Therefore, we need

   \begin{equation*} 5-m = 0, \ \ \text{ and } \ \ 3+2m+b = 0\text{.} \end{equation*}

   Solving, we find that \(m=5\) and \(b=13\text{.}\) With this choice, we see that

   \begin{equation*} \lim_{x\to \pm \infty} \left[\frac{5x^2-3x+1}{x+2} - (5x+13)\right]] = 0\text{,} \end{equation*}

   and so \(y=5x+13\) is the slant asymptote of \(f(x)\text{.}\)

2. \(f(x)=\dfrac{x^2}{x-1}\)

   Answer
   \(y=x+1\)
   Solution

   We wish to find a line \(y=mx+b\) such that

   \begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (mx+b)\right] = 0\text{.} \end{equation*}

   Therefore, we write

   \begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{x^2-mx^2+(b-m)x-b}{x-1} \\ \amp = \lim_{x\to \pm \infty} \frac{x-mx+b-m-b/x}{1-1/x}. \end{split} \end{equation*}

   In order for this limit to equal zero, we require \(m=1\) and \(b=1\text{.}\) With this choice, we find

   \begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2}{x-1} - (x+1)\right] = 0\text{,} \end{equation*}

   and so the line \(y=x+1\) is the slant asymptote of \(f(x)\text{.}\)

3. \(f(x)=\dfrac{x^2+3x+2}{x-1}\)

   Answer
   \(y=x+4\)
   Solution

   Consider

   \begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{(x^2+3x+2)-(x-1)(mx+b)}{x-1}\right] \amp = \lim_{x\to \pm \infty} \frac{(x^2+3x+2)-mx^2-(b-m)x+b}{x-1}\\ \amp = \lim_{x\to \pm \infty} \frac{x+3+2/x-mx-(b-m)+b/x}{1-1/x}. \end{split} \end{equation*}

   Therefore, if we take \(m=1\) and \(b=4\text{,}\) we see that

   \begin{equation*} \lim_{x\to \pm \infty} \frac{x+3+2/x-x-3-1/x}{1-1/x} = \lim_{x\to\pm \infty} \frac{2/x-1/x}{1-1/x} = 0\text{.} \end{equation*}

   Hence, the line \(y=x+4\) is the slant asymptote of \(f(x)\text{.}\)

4. \(f(x)=\dfrac{x^2-5x+4}{x-3}\)

   Answer
   \(y=x-1\)
   Solution

   We wish to find a line \(y=mx+b\) such that

   \begin{equation*} \lim_{x\to \pm \infty} \left[\frac{x^2-5x+4}{x-3} - (mx+b)\right] = 0\text{.} \end{equation*}

   Therefore, we write

   \begin{equation*} \begin{split} \lim_{x\to \pm \infty} \left[\frac{x^2-5x+4}{x-3} - (mx+b)\right] \amp = \lim_{x\to \pm \infty} \frac{x^2-5x+4 - mx^2 + (3m-b)x - 3b}{x-3} \\ \amp = \lim_{x\to \pm \infty} \frac{x-5+4/x - mx +(3m-b) - 3b/x}{1-3/x}. \end{split} \end{equation*}

   So if we take \(m=1\) and \(b=-1\text{,}\) we see that

   \begin{equation*} \begin{split} \lim_{x\to \pm \infty} \frac{x-5+4/x - mx +(3m-b) - 3b/x}{1-3/x} \amp = \lim_{x\to \pm \infty} \frac{x-5+4/x-x+5+3/x}{1-3/x}\\ \amp = 0. \end{split} \end{equation*}

   Therefore, the line \(y=x-1\) is the slant asymptote of \(f(x)\text{.}\)

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. \(\ds y=x^5-5x^4+5x^3\)

   Answer
   We wish to sketch the graph of \(y=f(x)=x^5-5x^4+5x^3\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

2. \(\ds y=x^3-3x^2-9x+5\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x^3-3x^2-9x+5\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

3. \(\ds y=(x-1)^2(x+3)^{2/3}\)

   Answer

   We wish to sketch the graph of \(y=f(x)=(x-1)^2(x+3)^{2/3}\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

4. \(\ds x^2+x^2y^2=a^2y^2\text{,}\) \(a>0\text{.}\)

   Answer

   We wish to sketch the graph of \(x^2+x^2y^2=a^2y^2\text{,}\) \(a>0\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

5. \(\ds y= x^5 - x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x^5-x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

6. \(\ds y=x(x^2+1)\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x(x^2+1)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

7. \(\ds y=x^3+6x^2 + 9x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x^3+6x^2+9x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. \(\ds y = 4x+\sqrt{1-x}\)

   Answer
   
   Solution

   We follow the Curve Sketching Guideline to sketch the graph of \(y=f(x)=4x+\sqrt{1-x}\text{:}\)

   1. The domain of \(y\) is all \(x\) such that \(1-x \geq 0 \implies x \leq 1\text{.}\) We further notice that \(f(1) = 4\text{.}\)

   2. The function has a \(y\)-intercept at

      \begin{equation*} y = f(0) = 1\text{.} \end{equation*}

      To find any \(x\)-intercepts, we set \(f(x)=0\text{:}\)

      \begin{equation*} \begin{split} 0 \amp = 4x + \sqrt{1-x} \\ -4x \amp = \sqrt{1-x} \\ 16x^2 + x -1 \amp = 0 \ \ (x \lt 0)\\ x\amp = -\frac{1}{32}-\frac{\sqrt{65}}{32} \approx -0.28 \end{split} \end{equation*}

      And so \(f\) has one \(x\)-intercept. We provide a sketch of our work so far below:

      

   3. There is no apparent symmetry.

   4. There are no horizontal or vertical asymptotes.

   5. We compute

      \begin{equation*} f'(x) = 4 - \frac{1}{2\sqrt{1-x}}\text{.} \end{equation*}

      Therefore, \(f(x)\) has critical points at \(x=1\) (where \(f'(x)\) undefined) and where

      \begin{equation*} f'(x) = 0 \implies \sqrt{1-x} = \frac{1}{8} \implies x = \frac{63}{64} \approx 0.98\text{.} \end{equation*}

      Since

      \begin{equation*} f'(0) = 4 - \frac{1}{2} > 0 \ \ \text{ and } \ \ f'(0.99) = -1 \lt 0\text{,} \end{equation*}

      we conclude that \(f\) is increasing on the interval \(\left(-\infty, \frac{63}{64}\right)\) and decreasing on \(\left(\frac{63}{64}, 1\right)\text{.}\) By the First Derivative Test, the point \(\left(\frac{63}{54}, \frac{65}{16}\right)\) is a local maximum. The point \((1,4)\) is not a local extrema.

   6. The second derivative of \(f\) is

      \begin{equation*} f''(x) = \frac{1}{4(1-x)^{3/2}}\text{.} \end{equation*}

      Therefore, \(f''(x)\) DNE at \(x=1\) and \(f''(x)=0\) has no solutions. Since

      \begin{equation*} f''(0) = -\frac{1}{2} \lt 0\text{,} \end{equation*}

      we conclude that \(y\) is concave down on \((-\infty, 1)\text{.}\) A final sketch is provided below.

      

2. \(\ds y = (x+1)/\sqrt{5x^2 + 35}\)

   Answer

   We wish to sketch the graph of \(y=f(x)=(x+1)/\sqrt{5x^2+35}\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

3. \(\ds y = x+ 1/x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x+1/x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

4. \(\ds y = x^2+ 1/x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x^2+1/x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

5. \(\ds y = (x+5)^{1/4}\)

   Answer

   We wish to sketch the graph of \(y=f(x)=(x+5)^{1/4}\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

6. \(\ds y=x/(x^2-9)\)

   Answer
   
   Solution

   We wish to sketch the graph of \(y=f(x)=\dfrac{x}{x^2-9}\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We see that \(f(x)\) is undefined at \(x = \pm 3\) and that \(f(0)= 0\text{.}\) Next, we look for asymptotes of \(f\text{.}\) Since

   \begin{equation*} \begin{array}{lll} \lim\limits_{x\to\infty} f(x) = 0 \amp \lim\limits_{x\to 3^+} f(x) = +\infty \amp \lim\limits_{x\to -3^+} f(x) = +\infty \\ \lim\limits_{x\to -\infty}f(x) = 0 \amp \lim\limits_{x \to 3^-} f(x) = -\infty \amp \lim\limits_{x\to -3^-} f(x) = -\infty, \end{array} \end{equation*}

   we see that \(y=0\) is a horizontal asymptote and \(x=\pm 3\) are vertical asymptotes:

   

   Next, compute

   \begin{equation*} f'(x) = -\frac{x^2+9}{(x^2-9)^2}\text{.} \end{equation*}

   Since \(f'(x)=0\) has no solutions, and \(f'(x)\) is defined for all \(x \neq \pm 3\text{,}\) we conclude that \(f\) has no relative extrema. Additionally, \(f'(0) \lt 0\) and so \(f\) is decreasing on its entire domain. We now investigate the concavity of \(f\text{.}\)

   \begin{equation*} f''(x) = \frac{2x(x^2+27)}{(x^2-9)^3}\text{,} \end{equation*}

   therefore \(f''(x)=0\) at \(x=0\text{,}\) and \(f''(x)\) is defined for all \(x \neq \pm 3\text{.}\) By picking appropriate test points, we find that \(f\) is concave down on \((-\infty,-3)\cup(0,3)\) and concave up on \((-3,0)\cup(3,\infty)\text{.}\) Thus, \(x=0\) is an inflection point of \(f\text{.}\) We summarize these findings in the following diagram.

   

   We arrive at the following graph of \(f(x)\text{.}\)

   

7. \(\ds y=x^2/(x^2+9)\)

   Answer

   We wish to sketch the graph of \(y=f(x)=x/(x^2+9)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

8. \(\ds y=2\sqrt{x} - x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=2\sqrt{x}-x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

9. \(\ds y=(x-1)/(x^2)\)

   Answer

   We wish to sketch the graph of \(y=f(x)=(x-1)/(x^2)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

Follow the Curve Sketching Guideline provided in this section to sketch the graphs of the following functions.

1. \(\ds y=xe^x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=xe^x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

2. \(\ds y=(e^x+e^{-x})/2\)

   Answer

   We wish to sketch the graph of \(y=f(x)=(e^x+e^{-x})/2\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

3. \(\ds y=e^{-x}\cos x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=e^{-x}\cos x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

4. \(\ds y=e^x-\sin x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=e^{x} - \sin x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

5. \(\ds y=e^x/x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=e^x/x\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

6. \(\ds y = \tan^2 x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=\tan^2(x)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

7. \(\ds y =\cos^2 x - \sin^2 x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=\cos^2(x)-\sin^2(x)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

8. \(\ds y = \sin^3 x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=\sin^3(x)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

9. \(\ds y = 6x + \sin 3x\)

   Answer

   We wish to sketch the graph of \(y=f(x)=6x+\sin(3x)\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

   

10. \(\ds y=3\sin(x) - \sin^3(x)\text{,}\) for \(x\in[0,2\pi]\)

    Answer

    We wish to sketch the graph of \(y=f(x)=3\sin(x)-\sin^3(x)\) for \(x\in [0,2\pi]\text{.}\) Therefore, we use the curve sketching guidelines provided in the course notes. We arrive at the following graph of \(f(x)\text{.}\)

    

Sketch the graph of \(y=f(x)\) using the following information:

Domain: \((-\infty,\infty)\)

\(y\)-intercept: 2

Asymptotes: none

Increasing on: \((-\infty,0)\cup(4,\infty)\)

Decreasing on: \((0,4)\)

Relative Extrema: relative max at \((0,2)\text{,}\) relative min at \((4,-30)\)

Concavity: Downward on \((-\infty,2)\text{,}\) upward on \((2,\infty)\)

Inflection point: \((2,-14)\)

We are given three points on the graph of \(f\) which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point \((0,2)\) is a relative maximum and \((4,-30)\) is a relative minimum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of \(f\) in the following diagram:

Using this diagram, we can piece together the final graph:

Sketch the graph of \(y=f(x)\) using the following information:

Domain: \((-\infty,\infty)\)

\(y\)-intercept: 0

\(x\)-intercepts: 0,2

Asymptotes: none

Increasing on: \((1.5,\infty)\)

Decreasing on: \((-\infty,0)\cup(0,3/2)\)

Relative Extrema: relative min at \((3/2,-27/80)\)

Concavity: Downward on \((0,1)\text{,}\) upward on \((-\infty,0)\cup(1,\infty)\)

Inflection points: \((0,0)\) and \((1,-1/5)\)

We are given four points on the graph of \(f\) which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point \((3/2,-27/80)\) is a relative minimum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of \(f\) in the following diagrams:

Using this diagram, we can piece together the final graph:

Sketch the graph of \(y=f(x)\) using the following information:

Domain: \((-\infty,0)\cup(0,\infty)\)

\(x\)-intercept: 1

Asymptotes: \(x\)-axis and \(y\)-axis

Increasing on: \((0,2)\)

Decreasing on: \((-\infty,0)\cup(2,\infty)\)

Relative Extrema: relative max at \((2,1/4)\)

Concavity: Downward on \((-\infty,0)\cup(0,3)\text{,}\) upward on \((3,\infty)\)

Inflection point: \((3,2/9)\)

We are given three points on the graph of \(f\) which we can use to initialize our set of axes. Plot,

Additionally, we are told that the point \((2,1/4)\) is a relative maximum. We can add this information to our sketch as follows:

Next, we summarize the information on the shape of \(f\) in the following diagrams:

Using this diagram, we can piece together the final graph:

Sketch the graph of \(y=f(x)\) using the following information:

Domain: \((-\infty,\infty)\)

Symmetry: Odd \(x\)-intercepts: \(\pm 5\sqrt{5}\text{,}\)0

Asymptotes: none

Increasing on: \((-\infty,-2.15)\cup(2.15,\infty)\)

Decreasing on: \((-2.15,2.15)\)

Relative Extrema: relative max at \((-2.15,4.3)\text{,}\) relative min at \((2.15,-4.3)\)

Concavity: Upward on \((0,\infty)\text{,}\) Downward on \((-\infty, 0)\)

We wish to plot the graph of \(y=f(x) = x-5x^{1/3}\text{.}\) We first plot the given points:

Next, we use the fact that \(y\) has relative extrema at \(x = \pm 2.15\text{:}\)

Finally, we connect the points making sure that the graph is symmetric about the origin (odd-symmetry):