Calc2.46

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Find the absolute minimum and maximum on [-1,5] of the function f(x)=(1-x)e^{x}.

It is known that the only place an absolute minimum or maximum can occur on an interval is at one of the endpoints of the interval or at a critical point inside the interval. A critical point is a point where the function is defined and its derivative is either 0 or undefined. So start by finding the derivative.

f'(x)=-e^{x}+(1-x)e^{x}=-xe^{x}\,

This derivative is defined everywhere so the only possible critical numbers will be where the derivative is equal to 0.

-xe^{x}=0 can only be true if x=0 or e^{x}=0 but e^{x} is always positive so it is never equal to 0. Thus the only critical point comes when x=0. As stated before, the absolute max and min can only occur at an endpoint or at a critical number. So our only choices for x here are -1, 0, or 5. If these are our only choices, our easiest plan is to just plug in the three values and see which gives the highest value and which gives the lowest value.

f(-1)={\frac  {2}{e}}\,<1

f(0)=1\,

f(5)=-4e^{5}\,

Thus, the absolute minimum comes at the point (5,-4e^{5}) and the absolute maximum comes at the point (0,1).


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