Computations for graphing functions

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We will give some general guidelines for sketching the plot of a function.

Let’s get to the point. Here we use all of the tools we know to sketch the graph of $y=f(x)$ :

Find the $y$ -intercept, this is the point $(0,f(0))$ . Place this point on your graph.
Find any vertical asymptotes, these are points $x=a$ where $f(x)$ goes to infinity as $x$ goes to $a$ (from the right, left, or both).
If possible, find the $x$ -intercepts, the points where $f(x) = 0$ . Place these points on your graph.
Analyze end behavior: as $x \to \pm \infty$ , what happens to the graph of $f$ ? Does it have horizontal asymptotes, increase or decrease without bound, or have some other kind of behavior?
Compute $f'$ and $f''$ .
Find the critical points (the points where $f'(x) = 0$ or $f'(x)$ is undefined).
Use either the first or second derivative test to identify local extrema and/or find the intervals where your function is increasing/decreasing.
Find the candidates for inflection points, the points where $f''(x) = 0$ or $f''(x)$ is undefined.
Identify inflection points and concavity.
Determine an interval that shows all relevant behavior.

At this point you should be able to sketch the plot of your function.

Sketch the graph of the function defined by $f(x)=2x^3-3x^2-12x$ .

The $y$ -intercept is $(0,\answer [given]{0})$ . Place this point on your plot.

Which of the following are vertical asymptotes? Select all that apply.

$x=0$ $x=1$ $x=-1$ $x=\sqrt {2}$ There are no vertical asymptotes

In this case, $f(x) =2x^3-3x^2-12x$ , we can find the $x$ -intercepts. There are three $x$ intercepts. Call them $a$ , $b$ , and $c$ , and order them such that $a<b<c$ . Then

$\begin{align*} a &= \answer[given]{\frac{3-\sqrt{105}}{4}},\\ b &= \answer[given]{0}, \\ c &= \answer[given]{\frac{3+\sqrt{105}}{4}}. \end{align*}$

Which of the following best describes the end behavior of $f$ as $x \to \infty$ ?

$f$ increases without bound. $f$ decreases without bound. $f$ has a horizontal asymptote. $f$ has some other behavior at $\infty$ .

Which of the following best describes the end behavior of $f$ as $x \to -\infty$ ?

$f$ increases without bound. $f$ decreases without bound. $f$ has a horizontal asymptote. $f$ has some other behavior at $\infty$ .

Compute $f'(x)$ and $f''(x)$ , $f'(x) = \answer [given]{6x^2 -6x -12}=6(\answer {x^2-x-2})=6(x+1)(\answer {x-2}); \hspace {0.2in} f''(x) = \answer [given]{12x-6}=12(\answer {x-\frac {1}{2}}).$

The critical points are where $f'(x) = 0$ , thus we need to solve $6(x+1)(x-2) = 0$ for $x$ . This equation has two solutions. If we call them $a$ and $b$ , with $a<b$ , then what are $a$ and $b$ ? $a = \answer [given]{-1}\qquad \text {and}\qquad b = \answer [given]{2}.$

Mark the critical points $x=2$ and $x=-1$ on your plot.

Since $f'(x)=6(x+1)(x-2)$ , and both factors $(x+1)$ and $(x-2)$ are negative for $x<-1$ , it follows that $f'(x)>0$ , and , therefore, our function is increasing decreasing on $(-\infty ,-1)$ .
Similarly, since for $-1<x<2$ , $(x+1)>0$ and $(x-2)<0$ , the derivative is negative there, and , therefore, our function is increasing decreasing on $(-1,2)$ .
And, for $x>2$ , both factors $(x+1)$ and $x-2)$ are positive, and , therefore, our function is increasing decreasing on $(2,\infty )$ .

Hence $x=-1$ , corresponding to the point $(-1,7)$ is a local and $x=2$ , corresponding to the point $(2,-20)$ is local of $f(x)$ . Identify this on your plot.

The candidates for the inflection points are where $f''(x) = 0$ , thus we need to solve $12(x-\frac {1}{2})=0$ for $x$ .

The solution to this is $x = \answer {\frac {1}{2}}$ .

This is only a possible inflection point, since the concavity needs to change to make it a true inflection point.

We have that $f''(x)<0$ for $x<\frac {1}{2}$ , therefore $f$ is concave on $(-\infty ,\frac {1}{2})$ .

Similarly, $f''(x)>0$ for $x>\frac {1}{2}$ , therefore $f$ is concave on $(\frac {1}{2},\infty )$ .

So this point is is not a point of inflection.

Since all of this behavior as described above occurs on the interval $[-2,4]$ , we now have a complete sketch of $f(x)$ on this interval, see the figure below.

Sketch the plot of $f(x) = \begin {cases} xe^x+2 &\text {if $x<0$} \\ x^4-x^2+3 &\text {if $x \geq 0$}. \end {cases}$ NOTE: $\lim _{x \to -\infty }xe^x=0$ . You will need this limit when answering some questions below.
Try this on your own first, then either check with a friend, a graphing calculator (like Desmos), or check the online version.

Since this function is piecewise defined, we will analyze the cases $(-\infty ,0)$ and $[0,\infty )$ separately.

Because $f$ is piecewise defined, and potentially discontinuous at $0$ , it is important to understand the behavior of $f$ near $x = 0$ .

$\lim _{x \to 0^-} f(x) = \answer [given]{2}$

$\lim _{x \to 0^+} f(x) = \answer [given]{3}$

Moreover,

$f(0) = \answer [given]{3}$

Record this information on our graph with filled and unfilled circles.

Which of the following are vertical asymptotes on $(-\infty ,0)$ ? Select all that apply.

$x=0$ $x=1$ $x=-1$ $x=\sqrt {2}$ There are no vertical asymptotes

Which of the following are vertical asymptotes on $(0,\infty )$ ? Select all that apply.

$x=0$ $x=1$ $x=-1$ $x=\sqrt {2}$ There are no vertical asymptotes

Which of the following best describes the end behavior of $f$ as $x \to \infty$ ?

$f$ increases without bound. $f$ decreases without bound. $f$ has a horizontal asymptote of $y=2$ . $f$ has some other behavior at $\infty$ .

Which of the following best describes the end behavior of $f$ as $x \to -\infty$ ?

$f$ increases without bound. $f$ decreases without bound. $f$ has a horizontal asymptote of $y = 2$ . $f$ has some other behavior at $\infty$ .

We mark the location of the horizontal asymptote:

The derivative of $f$ on $(-\infty , 0)$ is $f'(x)= \answer [given]{xe^x+e^x}=e^x(\answer [given]{x+1})$

The derivative of $f$ on $(0, \infty )$ is $f'(x)= \answer [given]{4x^3-2x}=4x(\answer [given]{x^2-\frac {1}{2}})$

The critical points are where $f'(x) = 0$ or does not exist. $0$ is a critical point, since we have already seen it is a point of discontinuity for $f$ , and thus $f'(0)$ does not exist there.

On $(-\infty , 0)$ , $f$ has a critical point at $x = \answer [given]{-1}$

On $(0,\infty )$ , $f$ has a critical point at $x = \answer [given]{\frac {1}{\sqrt {2}}}$

Mark the critical points $x=-1$ and $x=\frac {1}{\sqrt {2}}$ on your plot.

Using the first derivative, we can see that

On $(-\infty , -1)$ , $f'(x)=e^x(x+1)$ and has the same sign as the factor $(x+1)$ , which is negative. Therefore, $f$ is on $(-\infty , -1)$ .

On $(-1, 0)$ , $f'(x)$ has the same sign as the factor $(x+1)$ , which is positive. Therefore, on $(-1, 0)$ .

On $(0, \frac {1}{\sqrt {2}})$ , since $f'(\frac {1}{4})= \frac {1}{16}-\frac {1}{2}<0$ , $f$ is increasing decreasing .

On $(\frac {1}{\sqrt {2}}, \infty )$ , since $f'(1)= 4(1-\frac {1}{2})>0$ , $f$ is increasing decreasing .

The second derivative of $f$ on $(-\infty , 0)$ is $f''(x)= \answer [given]{xe^x+2e^x}=e^x(\answer [given]{x+2})$

The second derivative of $f$ on $(0, \infty )$ is $f''(x)= \answer [given]{12x^2-2}=12(\answer [given]{x^2-\frac {1}{6}})$ The candidates for the inflection points are where $f''(x) = 0$ .

On $(-\infty , 0)$ , $f''$ has one zero, namely $x = \answer [given]{-2}$ . The sign of $f''$ changes from positive to negative negative to positive] through this point.

On $(0, \infty )$ , $f''$ has one zero, namely $x = \answer [given]{\frac {1}{\sqrt {6}}}$ . The sign of $f''$ changes from positive to negative negative to positive] through this point.

Since all of this behavior as described above occurs on the interval $[-5,2]$ , we now have a complete sketch of $y=f(x)$ on this interval, see the figure below.