## Calculus: Early Transcendentals 8th Edition

The absolute maxima are $f(2) = 18$ and $f(5) = 18$ The absolute minimum is $f(0) = -2$ The point $f(2) = 18$ is a local maximum. The point $f(4) = 14$ is a local minimum.
$f(x) = x^3-9x^2+24x-2$ Note that this function is continuous and differentiable for all $x$. We can find the points where $f'(x) = 0$: $f'(x) = 3x^2-18x+24 = 0$ $x^2-6x+8 = 0$ $(x-2)(x-4) = 0$ $x = 2,4$ We can verify the values of the function when $f'(x) = 0$ and the endpoints of the interval $[0,5]$: $f(0) = (0)^3-9(0)^2+24(0)-2 = -2$ $f(2) = (2)^3-9(2)^2+24(2)-2 = 18$ $f(4) = (4)^3-9(4)^2+24(4)-2 = 14$ $f(5) = (5)^3-9(5)^2+24(5)-2 = 18$ The absolute maxima are $f(2) = 18$ and $f(5) = 18$ The absolute minimum is $f(0) = -2$ The point $f(2) = 18$ is a local maximum. The point $f(4) = 14$ is a local minimum.