Let $f$ be the function with derivative given by $f'(x)=x^2-\dfrac{2}{x}$. On which of the following intervals is $f$ decreasing?

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We need to find values of $x$ that result in $f'(x)$ being 0 or undefined. We can do this much easier if we rewrite the equation:

$$f'(x)=x^2-\frac{2}{x}$$ $$f'(x)=\frac{x^3}{x}-\frac{2}{x}$$ $$f'(x)=\frac{x^3-2}{x}$$

When the numerator is equal to zero, $f'(x)=0$. $x=0$ will also be a critical point since it makes the denominator 0.

$$x^3-2=0$$ $$x^3=2$$ $$x=\sqrt[3]{2}$$

The table below summarizes testing of the three intervals determined by the critical numbers.

$$\begin{array}{|c|c|c|c|} \hline \text{Interval} & -\infty \lt x \lt 0 & 0 \lt x \lt \sqrt[3]{2} & \sqrt[3]{2} \lt x \lt \infty \\ \hline \text{ Test Value} & x=-1 & x=1 & x= 2 \\ \hline \text{ Sign of }f'(x) & f'(-1) \gt 0 & f'(1) \lt 0 & f'(2) \gt 0 \\ \hline \text{Conclusion} & \text{Increasing} & \text{Decreasing} & \text{Increasing} \\ \hline \end{array}$$

$f$ is therefore decreasing on $\boxed{(0,\sqrt[3]{2}]}$.

Should increasing or decreasing intervals use brackets or parenthesis? It depends, though usually you won't have to worry about it since only one answer choice will have the correct values.

Check out this detailed post if you want to dive down the rabbit hole.