We have the curve y=x^2 - 2x + ln e^2. Now we need to find the extreme point of the curve. For this, we first find the first derivative of the curve.
y = x^2 - 2x + ln e^2
=> y = x^2 - 2x + 2
y' = 2x - 2
Equate this to 0
=> 2x - 2 = 0
=> 2x = 2
=> x = 1
At x = 1
y = 1^2 - 2*1 + 2
=> y = 1 - 2 + 2
=> y = 1
Also y'' = 2 which is positive at x = 1, therefore this is the point with the minimum value.
Here the curve has the minimum value.
We derive the extreme point of y = x^2 - 2x + ln e^2 as ( 1 , 1).
To find the extreme value of y = x^2 - 2x + ln e^2.
y(x) = x^2 -2x + 2, as lne^2 = 2.
y(x) = (x-1)^2 -1+2.
y(x) = (x-1)^2+1.
Since (x-1)^2 is a perfect square, (x-1)^2 > = 0. y(x) > = y(1) = 1. So y(x is least when x= 1.
y(x) = (x-1)^2-1 +lne^2 is > = 1 for all x.
Therefore y(x) = x^2-2x+ln e^2 = 1 is the minimum of y (x), when x= 1.
If we want to determine the local extreme of the behavior curve, we'll use the first derivative of the function (the vertex is a local extreme).
f'(x) = (x^2-2x+ln e^2)'
f'(x) = 2x - 2 + 0
f'(x) = 2x - 2
We'll determine the critical value of x, that is the root of the first derivative.
2x - 2 = 0
We'll add 2 both sides:
2x = 2
We'll divide by 2:
x = 2/2
x = 1
Now, we'll calculate the y coordinate of the local extreme:
f(1) = (1)^2 - 2 + 2ln e, but ln e = 1
f(1) = 1 - 2 + 2
We'll eliminate like terms:
f(1) = 1
The coordinates of the local extreme of the behavior curve are (1 ; 1).