Question: User: Graph of f Graph of f′ The graphs of the function $f$ and its derivative $f′$ are shown above for $−1 \leq x \leq 4$. (a) Find the average rate of change of $f$ over the interval $−1 \leq x \leq 4$. For how many values of $x$ in the interval $−1 \leq x \leq 4$ does the instantaneous rate of change of $f$ equal the average rate of change of $f$ over that interval? Note On your AP Exam, you will handwrite your responses to free-response questions in a test booklet (b) Write an equation for the line tangent to the graph of $f$ at $x = 1$. (c) For each of $\lim_{x\to 2} \frac{f(x)−f(2)}{x−2}$ and $\lim_{x\to 3} \frac{f(x)−f(3)}{x−3}$, find the value or give a reason why it does not exist. (d) Let $g$ be the function defined by $g(x) = e^x f(x)$. Find $g′(0)$. $f$ graph: left curve rises from about $(-1,-2)$ to a peak near $(0,3)$, then decreases to a minimum near $(2,1)$, then rises to about $(3,2)$ and falls to about $(4,1)$; position_hint $\in \{\text{top-left}, \text{center}\}$. $f′$ graph: a curve starts high near $x=-1$ and decreases crossing the $x$-axis near $x=0$, reaches a minimum near $(1,-2)$, then rises crossing the $x$-axis near $x=2$, becomes a horizontal segment at $y=1$ from $x=2$ to $x=3$ with an open circle at $x=3$, and a horizontal segment at $y=-1$ from $x=3$ to $x=4$ with an open circle at $x=3$; position_hint $\in \{\text{top-right}, \text{center}\}$.

1. **State the problem:** We are given graphs of $f$ and its derivative $f'$ on the interval $[-1,4]$ and asked to find: (a) The average rate of change of $f$ over $[-1,4]$ and the number of $x$ values where the instantaneous rate equals this average. (b) The equation of the tangent line to $f$ at $x=1$. (c) The values or existence of the limits $\lim_{x\to 2} \frac{f(x)-f(2)}{x-2}$ and $\lim_{x\to 3} \frac{f(x)-f(3)}{x-3}$. (d) The derivative $g'(0)$ where $g(x) = e^x f(x)$. 2. **Average rate of change formula:** $$\text{Average rate of change} = \frac{f(b)-f(a)}{b-a}$$ where $a=-1$ and $b=4$. 3. **Find $f(-1)$ and $f(4)$ from the graph:** - $f(-1) \approx -2$ - $f(4) \approx 1$ 4. **Calculate average rate of change:** $$\frac{f(4)-f(-1)}{4-(-1)} = \frac{1 - (-2)}{5} = \frac{3}{5} = 0.6$$ 5. **Number of $x$ where instantaneous rate equals average rate:** By the Mean Value Theorem, there is at least one $x$ where $f'(x) = 0.6$. From the $f'$ graph description: - $f'$ crosses $y=0.6$ twice: once between $x=2$ and $x=3$ (since $f'$ is 1 on $(2,3)$) and once between $x=3$ and $x=4$ (where $f'$ is -1, so no), and once before $x=0$ (since $f'$ crosses zero near $0$ and is high near $-1$). More precisely: - $f'$ is above 0.6 near $x=-1$ and decreases crossing zero near $x=0$. - $f'$ is 1 on $(2,3)$, so equals 0.6 there. - $f'$ is -1 on $(3,4)$, so no. Hence, $f'(x) = 0.6$ at exactly two points in $[-1,4]$. 6. **Equation of tangent line at $x=1$:** Formula: $$y = f(a) + f'(a)(x - a)$$ From graph: - $f(1)$ is near the minimum, approximately $f(1) \approx 1.5$ (estimate between 1 and 2). - $f'(1)$ is minimum near $-2$. So, $$y = 1.5 - 2(x - 1) = 1.5 - 2x + 2 = 3.5 - 2x$$ 7. **Limits at $x=2$ and $x=3$:** Recall, $$\lim_{x\to c} \frac{f(x)-f(c)}{x-c} = f'(c)$$ if $f$ is differentiable at $c$. - At $x=2$, $f'(2)$ is where $f'$ crosses zero near $x=2$, so $f'(2) = 0$. - At $x=3$, $f'$ has a jump discontinuity: it is 1 on $(2,3)$ but -1 on $(3,4)$ with open circles at $x=3$. So $f'(3)$ does not exist. Therefore: - $\lim_{x\to 2} \frac{f(x)-f(2)}{x-2} = 0$ - $\lim_{x\to 3} \frac{f(x)-f(3)}{x-3}$ does not exist. 8. **Find $g'(0)$ for $g(x) = e^x f(x)$:** Use product rule: $$g'(x) = e^x f(x) + e^x f'(x) = e^x (f(x) + f'(x))$$ At $x=0$: - $f(0) \approx 3$ (peak) - $f'(0) = 0$ (crosses zero near 0) So, $$g'(0) = e^0 (3 + 0) = 1 \times 3 = 3$$ **Final answers:** (a) Average rate of change = $0.6$; instantaneous rate equals average rate at 2 values of $x$. (b) Tangent line at $x=1$: $y = 3.5 - 2x$. (c) $\lim_{x\to 2} = 0$; $\lim_{x\to 3}$ does not exist. (d) $g'(0) = 3$.