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].

1 Answer
Nov 30, 2016

Answer:

Please see the explanation.

Explanation:

Here is the graph of the expression:
Desmos.com

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.