CALCULUS
posted by Brittany .
Consider the function f(x)=sqrt(2–5x^2), –5≤x≤1. The absolute minimum value is ____ and this occurs at x equals ______?
Found the max value to be 2 at x=0.

f'(x) = (1/2)(2  5x^2)^(1/2) )(10x)
= 5x(25x^2)^(1/2)
= 0 for a max/min
This has only one solution, when x = 0
so the min occurs when x = 0 and
f(0) = √(20) = √2 
4x2+24x+16y232y12=0 need to know how to write in the standard form for ellipse it is a precalculus