Answer to Question #155734 in Calculus for Tom Garland

Question

Given the function f(x)=-x^3+3x^2+8, determine the absolute minimum value of f on the closed interval [-2,4].

Answer 1

1) $f(-2)=28$ .

2) $f(4)=-8.$

3) Derivative of $f$ is $f'(x)=-3x^2+6x = -3x(x-2)$.

It is equal to zero at the points $x=0$ and $x=2$. So we can suspect the local minimum to be at these points.

$f(0)=8, f(2)=12.$

We can see, that the global minimum of $f$ on the interval [-2, 4] is -8 at the point $x=4.$