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(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

CALCULUS  jamie, Saturday, September 17, 2011 at 11:43pm
4x2+24x+16y232y12=0 need to know how to write in the standard form for ellipse it is a precalculus
Answer This Question
Related Questions
 calculus  Consider the function f(x)=(x^3)(e^9x), 2 is less than or equal to x...
 Calculus  If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is the...
 Calculus  If m ≤ f(x) ≤ M for a ≤ x ≤ b, where m is ...
 Calculus  Consider the function f(x)=xsqrt(36−x^2), −1≤x&#...
 New Age Math  Suppose that y≤5x,3x≤y and 14x+15y≤1 together ...
 CALCULUS  Consider the function f(x)=5â€“4x^2 , 3â‰¤xâ‰¤1 The absolute maximum...
 Calculus  Let f be a differentiable function defined on the closed interval [a,...
 Calculus  1.Determine the maximum value of the function f(x)= 5^x x^5 on the ...
 Calculus  2. Determine the minimum value of the function f(x)= e^(x)  2e^x on...
 Math  Suppose that y≤5x,3x≤y and 14x+15...
More Related Questions