f'(x) > 0 means f is increasing
f' = (1-2x^2) e^-x^2
since e^z > 0 for all z, f is increasing where 1-2x^2 > 0, or -1/√2 < x < 1/√2
everywhere else it's decreasing
since f'(1/√2) = 0, and f changes from increasing to decreasing there, iot's a max.
similarly for min.
Suppose f'(x) = xe^-x^2.
a) On what interval is f increasing? On what interval is f decreasing?
b) Does f have a maximum value? Minimum value?
So should I assume that f'(x) > 0 since e has to be a positive number?
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