Question

How do I find the numbers c that satisfies the Mean value theorem for \[f(x)={{e}^{-2x}}\] on the interval \[\left[ 0,3 \right]\]?

Answer 1

Hint: To solve the problems of mean value theorem, we first have to find the derivative of the function. Find the functional value at the end points of the given interval. Then use the mean value theorem that states \[\dfrac{f(b)-f(a)}{b-a}=f'(c)\]. Here, c is the point that satisfies the mean value theorem.

Complete step by step solution:
To solve the given problem, we first have to find the derivative of the function. Differentiating the function \[f(x)={{e}^{-2x}}\] we get
\[\dfrac{d\left( f(x) \right)}{dx}=-2{{e}^{-2x}}\]
Let there be a point c such that it satisfies the mean value theorem. The mean value theorem for the given function \[f(x)={{e}^{-2x}}\] in the range \[\left[ 0,3 \right]\] can be written as
\[\dfrac{f(3)-f(0)}{3-0}=f'(x)\]
We need to find the value of \[f(3)\And f(0)\]. Substituting \[x=3\] in the equation of the given function, we get \[f(3)={{e}^{-2\times 3}}={{e}^{-6}}\]. Substituting \[x=0\] in the equation of the given function, we get \[f(0)={{e}^{-2\times 0}}=1\]. We have to find the value of \[f'(x)\] at \[x=c\], substituting this point in the first derivative of the function we get \[f'(c)=-2{{e}^{-2c}}\]. Using these values in the above expression of the mean value theorem, we get
\[\dfrac{{{e}^{-6}}-1}{3}=-2{{e}^{-2c}}\]
Dividing both sides of the above equation by \[-2\], and flipping the sides it can be expressed as
\[\Rightarrow {{e}^{-2c}}=\dfrac{{{e}^{-6}}-1}{-6}\]
Taking logarithm of both sides of equation, we get
\[\Rightarrow \ln \left( {{e}^{-2c}} \right)=\ln \left( \dfrac{{{e}^{-6}}-1}{-6} \right)\]
Simplifying the left-hand side of the above equation using the property \[\ln {{a}^{b}}=b\ln a\], we get
\[\Rightarrow -2c\ln \left( e \right)=\ln \left( \dfrac{{{e}^{-6}}-1}{-6} \right)\]
\[\Rightarrow -2c=\ln \left( \dfrac{{{e}^{-6}}-1}{-6} \right)\]
 Dividing both sides of the above equation by \[-2\], we get the value of c as
\[\Rightarrow c=-\dfrac{1}{2}\ln \left( \dfrac{{{e}^{-6}}-1}{-6} \right)\]
Evaluating the value of expression in the right-hand side, we get
\[c\approx 0.9\]

Note: Rolle’s theorem and Lagrange’s mean value theorem are one of the important theorems in calculus. Rolle’s theorem is just a special case of the Lagrange’s theorem, when the functional value at the endpoints of the given intervals is the same.