# Find Absolute Max and Min

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

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

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The usual way:

1. Find the critical points.
2. Evaluate $f$ at the critical points and the endpoints.
3. Compare the values.
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Awesome got it! – soniccool Jun 26 '12 at 4:53
@Arturo : 2. should be "Evaluate $f$ at the critical points that are contained in the given interval and also at the endpoints". Please let me know if I am wrong. – Rajesh D Jun 26 '12 at 5:03
@RajeshD: That is correct, though implied in the context. – Cameron Buie Jun 26 '12 at 5:05
@Rajesh: To me, "the critical points" refers to the critical points in the domain of the function. The domain here is the specified interval (what happens outside is irrelevant). And I said "and the endpoints", so I don't see how adding "also at" changes the meaning. – Arturo Magidin Jun 26 '12 at 5:05
@Arturo : Thanks for clarifying. I do not intend anything by "also", it just came as a part of the sentence. – Rajesh D Jun 26 '12 at 5:07

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.

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 $2$.
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.