Asked by Ruth Teshome
Find the absolute maximum and minimum values of f(x)=8x(1-x^2)^1/2
Answers
Answered by
Helper
The maximum is 6√43e−16≈0.888(it occurs at x=√43) and the minimum is −e−18≈−0.882(at x=−1).
Answered by
oobleck
Not so much help, @Helper.
f'(x) = 8/(1-x^2)^(3/2)
For an absolute max and min, you generally need an interval.
f(x) has asymptotes at x = ±1
The graph looks kind of like tan x
So there are no absolute extrema.
f'(x) = 8/(1-x^2)^(3/2)
For an absolute max and min, you generally need an interval.
f(x) has asymptotes at x = ±1
The graph looks kind of like tan x
So there are no absolute extrema.