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Worksheet The Derivative Function - Activity 1.4.5

Activity 11.

The graph of a function \(f\) is given in Figure 149.
A piecewise linear graph on an x–y coordinate plane. The graph starts near (–4.5, 4.5) and slopes downward to a low point at about (–1, 1). From there it rises sharply to a peak around (1, 5), then slopes downward again to roughly (4, 3). The axes are labeled x and y, with gridlines visible.
Figure 149. The graph of \(f(x)\text{.}\)

(a)

On Figure 149, carefully sketch the graph of \(f'\) from the graph of \(f\text{.}\)
Hint.
What is the slope of the function at every point?
Answer.
A graph on an x–y coordinate plane showing a piecewise linear blue function and its derivative marked in red. The blue graph descends from the left to a low point near (–1, 1), then rises sharply to a peak around (1, 5), and slopes downward again toward the right. Red dashed horizontal segments represent derivative values: a segment around y = –1 from x less than –1 with open circles at boundaries, a segment at y = 2 between –1 and 1 with open circles at x = –1 and x = 1, and a segment at y ≈ –0.5 for x greater than 1 with an open circle at x = 1. The axes include gridlines, labeled x and y.
Figure 150. The graph of \(f(x)\) (in blue) and \(f'(x)\) (in red,dashed).

(b)

Determine values of \(f(3)\) and \(f'(3)\) from the graph of \(f(x)\) (if they exist).
Answer.
\(f(3)=3\) and \(f'(3)=-0.5\text{.}\)

(c)

Do you think \(f'(1)\) exists? Defend your answer.
Answer.
It does not exist since the slope radically changes at \(x=1\text{.}\)

(d)

Compare the values of \(\displaystyle \lim_{h \rightarrow 0^-} \frac{f(1+h)-f(1)}{h}\) and \(\displaystyle \lim_{h \rightarrow 0^+} \frac{f(1+h)-f(1)}{h}\text{.}\) What does this tell us?
Hint.
To the left of \(x=1\text{,}\) what is the slope of the function? To the right of \(x=1\text{?,}\) what is this slope?
Answer.
\(\displaystyle \lim_{h \rightarrow 0^-} \frac{f(1+h)-f(1)}{h} = 2\) and \(\displaystyle \lim_{h \rightarrow 0^+} \frac{f(1+h)-f(1)}{h} = -0.5\text{.}\) Since these values are different, the two-sided limit \(\displaystyle \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}\) which represents \(f'(1)\) does not exist.

(e)

Write a piecewise function describing \(f'(x)\text{.}\)
Answer.
\(f'(x) = \begin{cases} -1 {\textrm{ if }} -5 < x < -1 \\ 2 {\textrm{ if }} -1 < x < 1 \\ -0.5 {\textrm{ if }} 1 < x < 5 \end{cases}\)

(f)

Does \(f\) or \(f'\) have a larger domain? Will this always be true? Why or why not?
Hint.
What is the domain of \(f'\text{?}\)
Answer.
The domain of \(f\) is \([-5,5]\text{.}\) The domain of \(f'\) is \([-5,-1) \cup (-1,1) \cup (1,5] \text{.}\) So \(f\) has the larger domain. This will always be true since the derivative does not exist if the function is not defined at a given point.
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