Read derivatives from a graph

(a) At $x = 0$. Just left of 0, $f$ is the segment from $(-3, 1)$ to $(0, 4)$, with slope $\frac{4 - 1}{0 - (-3)} = 1$. Just right of 0, $f$ is the segment from $(0, 4)$ to $(3, -2)$, with slope $\frac{-2 - 4}{3 - 0} = -2$. The graph has no break at 0, but the slopes from the two sides, 1 and -2, differ, so $f'(0)$ does not exist.

• Point 1 Names x = 0 and computes both one-sided slopes, 1 and -2.

(b) No. From the left, $f(x)$ approaches $-2$, the filled dot at $(3, -2)$, but from the right it approaches 1, the open circle at $(3, 1)$. So $f$ is not continuous at $x = 3$, and a function that is not continuous at a point is not differentiable there.

• Point 2 Answers no and gives the jump from -2 to 1 as the reason.

(c) The average rate of change is $\frac{f(5) - f(-3)}{5 - (-3)} = \frac{5 - 1}{8} = \frac{1}{2}$.

• Point 3 Uses the endpoint values f(5) = 5 and f(-3) = 1 and gets 1/2.

(d) By the product rule, $g'(x) = 1 \cdot f(x) + x f'(x)$. From the graph, $f(1) = 2$, and on the middle segment $f'(1) = -2$. So $g'(1) = 2 + 1 \cdot (-2) = 0$. Also $g(1) = 1 \cdot f(1) = 2$, so the tangent line is $y = 2 + 0(x - 1)$, which is $y = 2$.

• Point 4 Writes the product rule and reads f(1) = 2 and f'(1) = -2 to get 0.
• Point 5 Finds g(1) = 2 and writes the horizontal tangent line y = 2.

(e) By the quotient rule, $k'(x) = \frac{f'(x) \cdot x - f(x) \cdot 1}{x^2}$. On the last segment, $f(4) = 3$ and $f'(4) = \frac{5 - 1}{5 - 3} = 2$. So $k'(4) = \frac{2 \cdot 4 - 3}{16} = \frac{5}{16}$.

• Point 6 Writes the quotient rule in the right order, reads f(4) and f'(4), and gets 5/16.