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' =...
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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).