# Determine the extreme value of the function x^2-2x+2.

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### 2 Answers

The extreme value of a function is its value at the point where the value of the first derivative is equal to 0.

f(x) = x^2 - 2x + 2

=> f'(x) = 2x - 2

2x- 2 = 0

=> x = 1

At x = 1, f(x) = 1^2 - 2 + 2 = 1

**The extreme value of the given function is 1.**

To establish the extreme value of a function, we'll have to calculate the first derivative of the function.

Let's find the first derivative of the function f(x):

f'(x)=( x^2-2x+2)'=(x^2)'-(2x)'+(2)'

f'(x)=2x-2

Now we have to calculate the equation of the first derivative:

2x-2=0

We'll divide by 2:

x-1 = 0

x=1

That means that the function has an extreme point, for the critical value x=1.

f(1) = 1^2 - 2*1 + 2

f(1) = 1-2+2

We'll eliminate like terms:

f(1) = 1

**The extreme point of the function is a minimum point whose coordinates are: (1 ; 1).**