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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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3 Answers 3

up vote 1 down vote accepted

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

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.

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3  
All you quantum typists.... –  Eugene Jun 26 '12 at 4:40

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 $2$.

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