what is the absolute min and max value of f(x)=e^-x2 on the interval [-2,1]

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1 Answer

f(x)=e-x², f'(x)=-2xe-x²=0 at extremum, so x=0 is an extremum, f(0)=1. The asymptote is the x-axis.

When f(-2)=e-4 or 1/e4 and f(1)=e-1=1/e. So the absolute maximum is 1, and the minimum in the given range is 1/e4 (about 0.0183).

by Top Rated User (1.1m points)

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