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?

Summary Submit Skip Question

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

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.