Calculus 1 absolute minimum and maximum of a function?

Question

This is the question: Find the absolute maximum and absolute minimum values of f(x)=(x^2-16)/(x^2+16) on the interval [−5,5].

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Answer 1 · 2016-11-30T17:56:15

Please see the explanation.

Here is the graph of the expression:
![Desmos.com]( useruploads.socratic.org )

The graph shows that the maximums for the interval occur at the ends of the interval, $x = -5 and x = 5$ . There is an obvious minimum at $x = 0$

Let's see what The Calculus can tell us about it:

$f(x) = (x^2 - 16)/(x^2 + 16); -5 <=x <=5$

Add 0 to the numerator in the form + 16 - 16:

$f(x) = (x^2 + 16 - 16 - 16)/(x^2 + 16); -5 <=x <=5$

Separate into two fractions:

$f(x) = (x^2 + 16)/(x^2 + 16) - (16 + 16)/(x^2 + 16); -5 <=x <=5$

The first term becomes 1 and the numerator of the second term becomes -32:

$f(x) = 1 - 32/(x^2 + 16); -5 <=x <=5$

Compute the first derivative:

$f'(x) = (64x)/(x^2 + 16)^2; -5 <=x <=5$

This can only be 0 at $x = 0$

Perform the second derivative test:

$f''(x) = -64(3x^2 - 16)/(x^2 + 16)^3$

$f''(0) = 64$

This is a minimum.

The absolute maximum is:

$lim_(xtooo) (x^2 - 16)/(x^2 + 16)$

You can find that this is 1 by repeated application of L'Hopital's rule or by doing the quotient - remainder substitution that I did.