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).**