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

$f(x)$ has an absolute maximum value of $-3$ at $x=3$ and an absolute minimum value of $-19/3$ at $x=-2$.
$$f(x)=\frac{2}{3}x-5\hspace{1cm}-2\le x\le3$$ 1) Find the derivative $f'(x)$: $$f'(x)=\frac{2}{3}$$ Since the value of $f'(x)$ is constant and equals $2/3$, $f(x)$ has no critical points. 2) Evaluate $f(x)$ at the endpoints: - For $x=-2$: $$f(-2)=\frac{2}{3}\times(-2)-5=-\frac{4}{3}-5=-\frac{19}{3}$$ - For $x=3$: $$f(3)=\frac{2}{3}\times(3)-5=2-5=-3$$ This means the function $f(x)$ has an absolute maximum value of $-3$ at $x=3$ and an absolute minimum value of $-19/3$ at $x=-2$.