## Calculus 8th Edition

$$f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5],$$ The local maximum value is $$f(2)=18$$ and the local minimum value is $$f(4)=14$$ The values of $f$ at the endpoints of the interval are $$f(0)=-2 \text { and } f(5)=18$$ Thus, $$f(0)=-2$$ is the absolute minimum value, $$f(2)=f(5)=18$$ is the absolute maximum value.
$$f(x)=x^{3}-9 x^{2}+24 x-2, \quad [0,5],$$ Since $f$ is continuous on [0,5],we can use the Closed Interval Method $$f(x)=x^{3}-9 x^{2}+24 x-2$$ \begin{aligned} f^{\prime}(x) &=3 x^{2}-18 x+24 \\ & =3\left(x^{2}-6 x+8\right) \\ &=3(x-2)(x-4) \end{aligned} Since $f^{\prime}(x)$ exists for all $x$ the only critical numbers of $f$ occur when $f^{\prime}(x)=0$ that is, $$x=2 \text { or } x=4 .$$ For $0 \lt x \lt 2$ $$f^{\prime}(x) \gt 0$$ For $2 \lt x \lt 4$ $$f^{\prime}(x) \lt 0$$ For $4 \lt x \lt 5$ $$f^{\prime}(x) \gt 0$$ so $$f(2)=18$$ is a local maximum value and $$f(4)=14$$ is a local minimum value. The values of $f$ at the endpoints of the interval are $$f(0)=-2 \text { and } f(5)=18$$ Thus, $$f(0)=-2$$ is the absolute minimum value, $$f(2)=f(5)=18$$ is the absolute maximum value.