Question

How do you find local maximum value of f using the first and second derivative tests: `f(x)= x^2 + 8x -12`?

Answer 1

`f` has no local maxima. It has a local minimum when `x=-4`.

Explanation:

Find `f'(x)` and see when it `=0` or `"DNE"`. These are the critical numbers at which a local maximum could exist.

To determine whether or not the point is a local maximum, you could use either the first or second derivative tests.

For the first derivative test , create a sign chart where the important values are the critical numbers. If the signs change from positive to negative, the point is a local maximum.

For the second derivative test , plug the critical number(s) into the second derivative. If the value is negative, the function is concave down at that point meaning the point is a local maximum.

Now, let's do the work:

`f'(x)=2x+8`

`f'(x)=0` when `x=-4`.
`f'(x)` never `"DNE"`.

First derivative test:

`color(white)(xxxxxXXXxx)-4`
`larr----------rarr`
`color(white)(xxx)"NEGATIVE"color(white)(xxxxx)"POSITIVE"`

Since the sign of the first derivative goes from negative to positive, there is a local minimum when `x=-4`.

We can prove the same thing with the...

Second derivative test:

`f''(x)=2`

Thus, the second derivative is ALWAYS positive, and the function is always concave up, which results in a local minimum.

Therefore, the function has no local maxima.

We can check a graph, even though it is obvious that the graph will form a parabola facing up, which will have only a minimum at its vertex:
graph{x^2+8x-12 [-65, 66.67, -36.6, 29.23]}