find the relative extrema of the function f(x)= x^2-4x+5. ?

To find the relative extrema of the function f(x) = x^2 - 4x + 5, we first need to find the derivative of the function and set it equal to zero to find the critical points.

f'(x) = 2x - 4

Setting f'(x) = 0, we get:

2x - 4 = 0
2x = 4
x = 2

So, the critical point is x = 2.

To determine if this critical point is a relative maximum or minimum, we can use the second derivative test. Taking the second derivative of the function:

f''(x) = 2

Since the second derivative is positive, the critical point x = 2 is a relative minimum.

Therefore, the relative minimum of the function f(x) = x^2 - 4x + 5 is at x = 2.