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 \(\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}\)
\(\displaystyle f'(c) = \frac{2*cos(pi) + cos(2*pi) - 2*cos(0) - cos(0)}{pi-0}\)
\(\displaystyle f'(c) = \frac{-2 + 1 - 2 - 1}{pi-0} = \frac{-4}{pi}\)
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.
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 \(\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}\)
\(\displaystyle f'(c) = \frac{2*cos(pi) + cos(2*pi) - 2*cos(0) - cos(0)}{pi-0}\)
\(\displaystyle f'(c) = \frac{-2 + 1 - 2 - 1}{pi-0} = \frac{-4}{pi}\)
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.