S sully405 New member Joined May 19, 2005 Messages 5 Nov 2, 2005 #1 Find the absolute extrema of the following functions on the given intervals. a.) f(x)=(x^(2/3))(x^2-1) [-1,1] b.) f(x)=(x^(1/2))/(x^2 +1) [1/4,1/2] Find all values of c that satisfy the conclusion of the Mean Value Theorem of the following function: f(x)=x^(1/3)-x [-8,-1][/code]

Find the absolute extrema of the following functions on the given intervals. a.) f(x)=(x^(2/3))(x^2-1) [-1,1] b.) f(x)=(x^(1/2))/(x^2 +1) [1/4,1/2] Find all values of c that satisfy the conclusion of the Mean Value Theorem of the following function: f(x)=x^(1/3)-x [-8,-1][/code]

pka Elite Member Joined Jan 29, 2005 Messages 9,279 Nov 2, 2005 #2 First find the derivative: \(\displaystyle f(x) = \sqrt[3]{x} - x,\quad f'(x) = \frac{1}{{\sqrt[3]{{x^2 }}}} - 1\) Now you need to solve for c: \(\displaystyle f'(c) = \frac{{f( - 1) - f( - 8)}}{{\left( { - 1} \right) - \left( { - 8} \right)}} = \frac{{0 - \left( 6 \right)}}{7}\)

First find the derivative: \(\displaystyle f(x) = \sqrt[3]{x} - x,\quad f'(x) = \frac{1}{{\sqrt[3]{{x^2 }}}} - 1\) Now you need to solve for c: \(\displaystyle f'(c) = \frac{{f( - 1) - f( - 8)}}{{\left( { - 1} \right) - \left( { - 8} \right)}} = \frac{{0 - \left( 6 \right)}}{7}\)