Mean Value Theorem

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fcabanski

Junior Member
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Aug 17, 2013
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The problem states "Do not use a calculator." I know how to solve this with a calculator, or with a graphing utility such as a graphing calculator or wolfram. The problem is, I don't know how to solve it without a calculator.

Determine if the conditions for the MVT are met. If so, determine the value of c guaranteed by the Mean Value Theorem.

f(x) = 2cos(x) + cos(2x) on the interval [0,pi]

This function is continuous on the closed interval, and differentiable on the open interval. It meets the conditions for MVT.

MVT f′(c)=f(b)−f(a)b−a\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}

f′(c)=2∗cos(pi)+cos(2∗pi)−2∗cos(0)−cos(0)pi−0\displaystyle f'(c) = \frac{2*cos(pi) + cos(2*pi) - 2*cos(0) - cos(0)}{pi-0}

f′(c)=−2+1−2−1pi−0=−4pi\displaystyle f'(c) = \frac{-2 + 1 - 2 - 1}{pi-0} = \frac{-4}{pi}

Derivative of 2cos(c) + cos(2c) = -2sin(c) -2sin(2c) = -2 (sin(c) + sin(2c)) = -4/pi

sin(c) + sin(2c) = 2/pi

This is where I'm stumped. sin(2c) = 2coscsinc, but that doesn't seem to help.
 
Looks good to me!
Without using a numerical solution this is as far as you can go.

ANSWER:
c = g-1(2/pi)
where g(x)=sin(x) + sin(2x)
 
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