Calculus Question!

Question

Find the absolute maximum and absolute minimum values of f on the given interval.

f(x)= (x^2-64)/(x^2+64). [-64,64]

Answer 1

Expand the equation to get:

f(x) = x^4 - 3904

f"(x) = 4x^3

f"(x) = 12x^2

Set 1st derivative to zero to find critical values (inflection points, max, and min values)

4x^3 = 0

x = 0

There is only one critical value.

Now figure out if it is an inflection point, maximum, or minimum point

Use the 1st derivative test.

Check the slope using f'(x) for 2 test values less than 0 and greater than 0

When x = -1, f'(x) = -4 < 0 means slope is going down

When x = 1, f'(x) = 4 > 0 means slope is going up

So at x = 0, f(x) is a minimum value

(Note: the 2nd derivative = 0 in this case, so the 2nd derivative test doesn't help)

So at x=0, is it a relative or absolute minimum?

Check what is going on at the two endpoints, x = -64, and x = 64

At x=--64, f(x) = 16,773,312 (a huge positive number)

At x=-64, f(x) = 16,773,312 (same positive number)

These are both absolute maximums

The graph is concave up with:

absolute minimum at x = 0

2 absolute maximums at x = -64 and x = 64

Answer 2

There's a procedure for finding the absolute min and max of a continuous function on a closed interval. Maybe you've seen it before.

Find the critical points of the function within the interval. Then evaluate the function at each of these critical points and at the endpoints of the interval. The smallest of these values is the min, the largest is the max.