The minimum value of f(x) = 4x^2 + 4x + 1 is at the point x = a, which is the solution of f'(x) = 0. Also, f''(a) should be positive.

f'(x) = 8x + 4

f'(x) = 0

=> 8x + 4 = 0

=> x = -1/2

f''(x)...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

The minimum value of f(x) = 4x^2 + 4x + 1 is at the point x = a, which is the solution of f'(x) = 0. Also, f''(a) should be positive.

f'(x) = 8x + 4

f'(x) = 0

=> 8x + 4 = 0

=> x = -1/2

f''(x) = 4 which is positive for all values of x.

f(-1/2) = 4*(-1/2)^2 + 4*(-1/2) + 1 = 4*(1/4) - 2 + 1 = 0

**The minimum value of f(x) = 4x^2 + 4x + 1 is 0**