1. **Stating the problem:** We are given the derivatives of three functions: $$f'(x) = -3x + 4$$ $$g'(x) = 3(x - 1)^2 - 2$$ $$h'(x) = -7(x + 5)^2 - 1$$ We want to analyze these derivatives to understand the shapes of the original functions $f(x)$, $g(x)$, and $h(x)$. 2. **Understanding the problem:** The derivative $f'(x)$ tells us the slope of $f(x)$ at any point $x$. The sign and shape of $f'(x)$ help us determine where $f(x)$ is increasing or decreasing and the concavity. 3. **Analyzing $f'(x) = -3x + 4$:** - This is a linear function with slope $-3$. - It crosses zero at $x = \frac{4}{3}$. - For $x < \frac{4}{3}$, $f'(x) > 0$ so $f(x)$ is increasing. - For $x > \frac{4}{3}$, $f'(x) < 0$ so $f(x)$ is decreasing. - This means $f(x)$ has a maximum at $x = \frac{4}{3}$. 4. **Analyzing $g'(x) = 3(x - 1)^2 - 2$:** - Since $(x - 1)^2 \geq 0$, the minimum value of $g'(x)$ is $3 \cdot 0 - 2 = -2$. - $g'(x)$ is always greater than or equal to $-2$. - It equals zero when $3(x - 1)^2 - 2 = 0 \Rightarrow (x - 1)^2 = \frac{2}{3}$. - For $x$ near 1, $g'(x)$ is negative, but as $|x - 1|$ grows, $g'(x)$ becomes positive. - This suggests $g(x)$ decreases near $x=1$ and increases outside that interval. 5. **Analyzing $h'(x) = -7(x + 5)^2 - 1$:** - Since $(x + 5)^2 \geq 0$, $-7(x + 5)^2 \leq 0$. - Adding $-1$ makes $h'(x)$ always negative. - Therefore, $h'(x) < 0$ for all $x$. - This means $h(x)$ is strictly decreasing everywhere. 6. **Summary of shapes:** - $f(x)$ has a single maximum point. - $g(x)$ has a local minimum at $x=1$ and changes from decreasing to increasing. - $h(x)$ is always decreasing. 7. **Matching to graph options:** - The graph with a single peak corresponds to $f(x)$. - The graph with a valley corresponds to $g(x)$. - The graph that is always decreasing corresponds to $h(x)$. **Final answer:** The shapes correspond uniquely to $f(x)$, $g(x)$, and $h(x)$, so the correct choice is not "More than one option."