## University Calculus: Early Transcendentals (3rd Edition)

$h(x)$ has the absolute maximum value of $0$ at $x=0$ and the absolute minimum value of $-3$ at $x=-1$ and $x=1$ in the defined domain.
$$h(x)=-3x^{2/3}\hspace{1cm}-1\le x\le1$$ 1) Find the derivative $h'(x)$: $$h'(x)=-3\times\frac{2}{3}x^{-1/3}=-2x^{-1/3}$$ Here we have $h'(x)=0$ only when $x=0$. Therefore, $x=0$ is the only critical point of $h(x)$. 2) Evaluate $h'(x)$ at the critical points and endpoints: - For $x=-1$: $$h(-1)=-3(-1)^{2/3}=-3(1)=-3$$ - For $x=0$: $$h(0)=-3(0)^{2/3}=-3(0)=0$$ - For $x=1$: $$h(8)=-3(1)^{2/3}=-3(1)=-3$$ This means $h(x)$ has the absolute maximum value of $0$ at $x=0$ and the absolute minimum value of $-3$ at $x=-1$ and $x=1$ in the defined domain.