Posted by **Brittany** on Saturday, July 16, 2011 at 3:31am.

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.

- CALCULUS -
**Reiny**, Saturday, July 16, 2011 at 8:12am
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

- CALCULUS -
**jamie**, Saturday, September 17, 2011 at 11:43pm
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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