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