Question

Find the number “ c ” that satisfy the Mean Value Theorem (M.V.T) on the given

intervals.

(a) f(x)= e^-x , [0, 2]

(b) f(x)= x/(x+2), (1,

Answer 1

By mean value theorem,

$\text{ There exists} c\in [a,b].\\ f'(c)=\frac{f(b)-f(a)}{b-a}\\ a) f=e^{-x}\\ f'=-e^{-x}\\ \text{Then, by MVT}\\ f'(c)=\frac{f(2)-f(0)}{2-0}\\ -e^{-c}=\frac{e^{-2}-1}{2-0}\\ -e^{-c}=-0.43\\ -c=ln(0.43)\\ c=0.84\in [0,2].\\ b)\\ f=\frac{x}{x+2}\\ f'=\frac{2}{(x+2)^2}\\ \text{Then, by MVT}\\ f'(c)=\frac{f(2)-f(1)}{2-1}\\ \frac{2}{(c+2)^2}=\frac{\frac{1}{2} -\frac{1}{3}}{2-1}\\ \frac{2}{(c+2)^2}=\frac{1}{6}\\ (c+2)^2=12\\ c+2=2\sqrt3\\ c=2\sqrt3-2\\ c=1.46\in (1,2)$