locate the absolute extrema of the function f (x) = x^2 + 4x + 3 [-4,4]

Question

so far I've done = 2x + 4x + 3 = 0

2x + 4 = 0

and then do i plug the x and y value in the original equation? or the derivative equation?

Answer 1

Absolute extrema can either exists at endpoints or at critical points.

Find the critical point by setting the derivative of the function equal to 0 (or undefined).

f '(x) = 2x + 4

f '(c) = 2c + 4 = 0

2c = -4

c = -2, which exists on the interval [-4, 4]

Therefore, -2 is a critical value.

You're trying to find the highest and lowest values of the function so you would plug into the original function.

Get the y values of the endpoints and the critical value.

f(-4) = 3

f(4) = 35

f(-2) = -1.

The absolute max is the greatest of these: 35

The absolute min is the least of these: -1

Answer 2

f'(x) = 2x + 4

0 = 2x + 4

-4 = 2x

-2 = x