1. The problem asks us to identify which graph represents the derivative $f'(x)$ of the function $f(x)$ shown. 2. Recall that the derivative $f'(x)$ gives the slope of the tangent line to the curve $y=f(x)$ at each point $x$. 3. Important rules: - Where $f(x)$ has a local maximum or minimum, $f'(x)=0$ because the slope is zero. - Where $f(x)$ is increasing, $f'(x)$ is positive. - Where $f(x)$ is decreasing, $f'(x)$ is negative. 4. Analyze the given $f(x)$ curve: - It starts slightly below $y=1$ at $x=-2$ and rises to a maximum near $(0,3)$. - At $x=0$, $f(x)$ has a local maximum, so $f'(0)=0$. - After $x=0$, $f(x)$ falls steeply, crossing $y=0$ near $x=1$. 5. Therefore, $f'(x)$ should: - Be positive before $x=0$ (since $f(x)$ is increasing). - Equal zero at $x=0$ (local max). - Be negative after $x=0$ (since $f(x)$ is decreasing). 6. Check each option: - a) A straight line descending from about $(0,4)$ to $(2,-1)$: at $x=0$, $f'(0)=4$ (not zero), so no. - b) A straight line ascending from about $(-1,-2)$ to $(2,3)$: at $x=0$, $f'(0)$ is positive, no zero at max, so no. - c) A curve starting near $(-3,0)$, rising to about $(-1,3)$, then falling steeply crossing $y=0$ near $x=1$: this matches the behavior of $f'(x)$, with zero near $x=0$ and changing sign. - d) A straight line descending from about $(-1,2)$ to $(1,-2)$: at $x=0$, $f'(0)$ is positive, not zero, so no. 7. Conclusion: Option c) best represents $f'(x)$. Final answer: The curve in option c) represents $f'(x)$.