# Calculus 1 absolute minimum and maximum of a function?

## 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].

Nov 30, 2016

#### Explanation:

Here is the graph of the expression:

The graph shows that the maximums for the interval occur at the ends of the interval, $x = - 5 \mathmr{and} x = 5$. 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:

$f ' ' \left(x\right) = - 64 \frac{3 {x}^{2} - 16}{{x}^{2} + 16} ^ 3$

$f ' ' \left(0\right) = 64$

This is a minimum.

The absolute maximum is:

${\lim}_{x \to \infty} \frac{{x}^{2} - 16}{{x}^{2} + 16}$

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.