# CALCULUS

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.

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1. f'(x) = (1/2)(2 - 5x^2)^(-1/2) )(-10x)
= -5x(2-5x^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) = √(2-0) = √2

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posted by Reiny
2. 4x2+24x+16y2-32y-12=0 need to know how to write in the standard form for ellipse it is a precalculus

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posted by jamie

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