Find Absolute Max and Min

Question

How do i find the absolute maximum and absolute minimum values of f on this given interval.

$f(x) = 6x^3 − 9x^2 − 36x + 7, \ [−2, 3]$

Answer 1

The usual way:

1. Find the critical points.
2. Evaluate $f$ at the critical points and the endpoints.
3. Compare the values.

Answer 2

Find the critical points at which the derivative is zero.

Figure out if these critical points are local maximum/ local minimum.

Also, evaluate the function at the end points of the interval.

Now you should be able to find the absolute maximum and absolute minimum.

Move your mouse over the gray area for the answer.

We have $f(x) = 6x^3-9x^2 - 36x+7$. This implies that $$f'(x) = 18x^2 -18x -36 = 18 (x^2-x-2) = 18(x-2)(x+1)$$ Setting $f'(x) = 0$, we get the critical points as $x=-1,+2$. The functional value at these points is $f(2) = -53$ and $f(-1) = 28$. The functional value at the end points are $f(-2) = -5$ and $f(3) = -20$. Hence, the global maximum for $f(x)$ in the interval $[-2,3]$ is $28$ at $x=-1$ and the global minimum is $-53$ at $2$.

Answer 3

For differentiable functions on a closed interval, the absolute extrema will occur at critical points or at the endpoints. All you need to do is find the $x$ in the interval (if any) at which $f'(x)=0$ (or at which $f'(x)$ is undefined, in the general case), and check the values of $f(x)$ at those $x$-values and the endpoints.