# Determine the minimum value of the function f(x)=4x^2-8x+1.

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### 2 Answers

We have the function f(x) = 4x^2 - 8x + 1

f'(x) = 8x - 8

f''(x) = 8

For f'(x) = 0

=> 8x - 8 = 0

=> x = 1

At x = 1, f''(x) is positive. So the value of f(x) at x = 1 is the minimum.

f(1) = 4 - 8 + 1 = -3

**The minimum value of the function is -3**

The minimum of the function is reached for the critical value of the function. The critical value represents the root of the first derivative of the function.

We'll determine the 1st derivative, differentiating with respect to x:

f'(x) = 8x - 8

We'll cancel out f'(x):

f'(x) = 0

8x - 8 = 0

8x = 8

x = 1

We'll calculate the minimum of the function for the critical value x = 1.

f(1) = 4 - 8 + 1

f(1) = -3

**The minimum value of the function is represented by the pair:(1 ; -3).**