# What is the minimum value of f(x) = 4x^2 + 4x + 1

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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**

The function f(x) = 4x^2 + 4x + 1 can be written as follows:

f(x) = 4x^2 + 4x + 1

Use the relation (a+b)^2 = a^2 + b^2 + 2*a*b

= (2x + 1)^2

The value of a number that has been raised to the power 2 cannot go less than 0. The least value of (2x + 1)^2 is 0 at x = -1/2.

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