you are correct.
f' = 1 - x/√(1-x^2)
f'=0 at x = 1/√2
so there is a local max at x = 1/√2
Now check f(-1) and f(1) to see whether those values are greater or less than f(1/√2)
The absolute max is the greatest of these, and the absolute min is the least.
Find the absolute max and min values of the following functions on the given intervals.
f(x)= x+ sqrt(1-x^2) on [-1,1]
I found the derivative to be 1+1/2(1-x^2)^-1/2*(-2x), where do I go from there?
1 answer