# Math Insight

### The derivative, critical points, and graphing

Math 201, Spring 19
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Total points: 3
1. As a warm up, consider the function $f(x)=\left(x^{2} - 1\right) e^{x}$ that is graphed below.
Feedback from applet
Critical points:
Number of critical points:
Number of zeros:
Zeros:
1. Calculate the derivative of $f$.
$\diff{f}{x} =$
2. Calculate the critical points of $f$, the points where $\diff{f}{x}=0$ or $\diff{f}{x}$ does not exist.

Critical points:

Write the answers in increasing order, separated by commas.

Evaluate $f$ at each of those critical points.
Write your answers as ordered pairs of the form $(a,b)$, where $a$ is the critical point and $b$ is the value of $f$ at the critical point.
For first critical point:

For second critical point:

Plot critical points on the above graph, i.e., plot the points $(a,b)$ you just calculated.

3. The two critical points divide the number line into three intervals: one to the left of the critical points, one between the critical points, and one to the right of the critical points. What are these three intervals?

Left interval:

Middle interval:

Right interval:

4. On each of these intervals, $\diff{f}{x}$ does not change sign. Pick an auxiliary point in each interval, test the sign of $\diff{f}{x}$ at that point, and conclude whether $f$ is increasing or decreasing on that interval.

On left interval $f$ is

On middle interval $f$ is

On right interval $f$ is

5. What are the two roots of $f$ itself, i.e., at what points is $f(x)=0$?

Roots of $f$:

What is the value of $f$ at each of those roots?

Plot the roots of $f$ on the above graph.

(Usually, the next step is to sketch the graph of $f$ from this information, but in this warm-up problem, you have the graph of $f$.)

2. Let $f(x) = \frac{1}{4} \left(x - 4\right) \left(x - 1\right) \left(x + 4\right) = \frac{x^{3}}{4} - \frac{x^{2}}{4} - 4 x + 4$.
1. Calculate the derivative of $f$.
$f'(x)=$
2. Find the critical points of $f$, i.e., the points where $f'(x)=0$ or $f'(x)$ does not exist.

The critical points are $x=$
and $x=$
.
Enter in increasing order.

Calculate the value of $f$ at the critical points.
$f=$
and
at the first and second critical points, respectively.

3. Since $f$ is defined everywhere, the critical points are the only points where $f'(x)$ can change sign. The two critical points divide the number line into three intervals: one to the left of the critical points, one between the critical points, and one to the right of the critical points. What are these three intervals?

Left interval:

Middle interval:

Right interval:

4. On each of these intervals, $f'(x)$ does not change sign. Pick an auxiliary point in each interval, test the sign of $f'(x)$ at that point, and conclude whether $f$ is increasing or decreasing on that interval.

For the left interval, $f$ is

For the middle interval, $f$ is

For the right interval, $f$ is

5. What are the roots of $f$ itself, i.e., at what points is $f(x)=0$?

Roots of $f$: $x=$

Separate answers by commas.

6. Using this information, sketch the graph of $f(x)$.
Feedback from applet
Critical points:
Number of critical points:
Number of zeros:
Zeros:

3. Let $f(x) = \frac{4 x^{3}}{3} - 12 x^{2} + 36 x - 30$.
1. Calculate $f'(x) =$
2. Find the critical points of $f$.

$x=$

Enter answers in increasing order, separated by commas.

Find the value of $f$ at the critical points:

Enter answers in the same order as above, separated by commas.

3. What are the intervals over which $f'(x)$ cannot change sign, as determined by the critical points? Determine whether $f$ is increasing or decreasing on each interval.

Left interval:

Right interval:

In the left interval, $f$ is

In the right interval, $f$ is

4. Using this information and the value of $f(0) =$
, sketch the graph of $f(x)$.

4. Let $f(x) = \left\lvert{x + 1}\right\rvert$, which is the same as the function $$f(x) = \begin{cases} x+1 & \text{if x \gt -1}\\ -x-1 & \text{if x \lt -1}\\ 0 & \text{if x = -1}. \end{cases}$$
1. Calculate $f'(x)$. Be sure to note any points where the derivative does not exist.

$f'(x) =$
if $x >$

$f'(x) =$
if $x <$

$f'(x)$ does not exist if $x=$

2. Find the critical points of $f$.

Critical points: $x=$

Enter answers in increasing order, separated by commas.

3. Determine the intervals over which $f$ is increasing and decreasing.

Left interval: $f$ is
on the interval

Right interval: $f$ is
on the interval

4. Using this information and the value of $f(0) =$
, sketch the graph of $f(x)$.

5. Let $f(x) =e^{- \frac{1}{2} \left(x - 2\right)^{2}}$.
1. Calculate $f'(x)=$
2. Find the critical points of $f$.

Critical points: $x=$

Enter in increasing order, separated by commas.

Calculate the values of $f$ at the critical points:

Enter in same order as the critical points, separated by commas.

3. Determine the intervals over which $f$ is increasing and decreasing.

Left interval: $f$ is
on the interval

Right interval: $f$ is
on the interval

4. Using this information and the values of $f(10)$ and $f(-6)$, sketch the graph of the function.

$f(10) =$
, $f(-6)=$
.