If $g$ is a differentiable function such that $g(x)\gt 0$ for all real numbers $x$ and if $f'(x)=(x^2-9)g(x)$, which of the following is true?

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$$f'(x)=(x^2-9)g(x)$$ $$f'(x)=(x+3)(x-3)g(x)$$

Since $g(x)$ is always greater than 0, we can simply treat it as a positive constant such as $1$. The critical values are $-3$ and $3$.

$$f'(x)=(x+3)(x-3)(1)$$

Testing the intervals:

$$\begin{array}{|c|c|c|c|} \hline \text{Interval} & -\infty \lt x \lt -3 & -3 \lt x \lt 3 & 3 \lt x \lt \infty \\ \hline \text{ Test Value} & x=-4 & x=0 & x= 4 \\ \hline \text{ Sign of }f'(x) & f'(-4) \gt 0 & f'(0) \lt 0 & f'(4) \gt 0 \\ \hline \text{Conclusion} & \text{Increasing} & \text{Decreasing} & \text{Increasing} \\ \hline \end{array}$$

At $-3$, we have a relative maximum since the sign changes from positive to negative. At $3$, we have a relative minimum since the sign changes from negative to positive.