Math Insight

The derivative, critical points, and graphing

Math 1241, Fall 2019
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ID #:
Due date: Nov. 1, 2019, 11:59 p.m.
Table/group #:
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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.

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