Determine the minimum value of the function f(x)= e^(-x) - 2e^x on the interval -1≤x≤2

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rcmath | High School Teacher | (Level 1) Associate Educator

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Notice that since f' is always negative, then f is decreasing. That is why we chose x=2.
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rcmath | High School Teacher | (Level 1) Associate Educator

Posted on

Let's look at the first derivative of this function to see its concavity, and if it has a min or a max.

`f'(x)=-e^(-x)-2e^x =>`

`f'(x)=-1(e^(-x)+2e^x)`

Notice that the value between parenthesis is always positive, which means that f'(x) is always negative and none zero regardless of the value of x, hence no absolute minimun. But the minimum on the interval given will have to be at the point x=2.

So the minimum of the function over the given interval is:

`f(2)=e^(-2)-2e^2=1/e^2-2e^2=[1-2e^4]/e^2`

 

 

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