Question

Find $c$ so that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ if $f\left( x \right) = {e^x}$, $a = 0$ and $b = 1$.

Answer 1

Here, we need to find the value of $c$. We will find the derivative of the given function. Then, using the given information, we will simplify the given equation. We will take natural logarithms of both sides to simplify the equation further and get the value of $c$.

Complete step by step solution:
First, we will find the value of the function $f\left( x \right) = {e^x}$ at $x = 0$ and $x = 1$.
We will use the rule of logarithms $\ln {e^x} = x$.
Substituting $x = 0$ in the function, we get
$f\left( 0 \right) = {e^0}$
We know that any number raised to the power 0 is equal to 1.
Thus, we get
$\Rightarrow f\left( 0 \right) = 1$
Substituting $x = 1$ in the function $f\left( x \right) = {e^x}$, we get
$f\left( 1 \right) = {e^1}$
We know that any number raised to power 1 is the number itself.
Thus, we get
$\Rightarrow f\left( 1 \right) = e$
Now, we will find the value of the derivative of the function at $x = c$.
Differentiating the function $f\left( x \right) = {e^x}$ with respect to $x$, we get
$f'\left( x \right) = {e^x}$
Substituting $x = c$ in the equation, we get
$\Rightarrow f'\left( c \right) = {e^c}$
Finally, we can begin simplifying the given equation to find the value of $c$.
Substituting $a = 0$ and $b = 1$ in the equation $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$, we get
$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}}$
Substituting $f'\left( c \right) = {e^c}$, $f\left( 1 \right) = e$ and $f\left( 0 \right) = 1$ in the equation, we get
$\begin{array}{l} \Rightarrow {e^c} = \dfrac{{e - 1}}{1}\\\ \Rightarrow {e^c} = e - 1\end{array}$
Taking natural logarithms on both sides, we get
$\Rightarrow \ln {e^c} = \ln \left( {e - 1} \right)$
We know that $\ln {e^x} = x$.
Therefore, we get
$\ln {e^c} = c$
Substituting $\ln {e^c} = c$ in the equation $\ln {e^c} = \ln \left( {e - 1} \right)$, we get
$c = \ln \left( {e - 1} \right)$

$\therefore$ The value of $c$ is $\ln \left( {e - 1} \right)$.

Note:
We need to remember that any number raised to the power 0 is equal to 1. A common mistake we might commit is to $f\left( b \right) - f\left( a \right)$ as ${e^1} - {e^0} = e - 0 = e$, it will lead us to wrong answer. This question verifies a fundamental result in Calculus, called the Mean Value Theorem. According to the Mean Value Theorem, if $f:\left[ {a,b} \right] \to {\bf{R}}$ is a continuous function on the interval $\left[ {a,b} \right]$ and a differentiable function on the interval $\left( {a,b} \right)$, then there exists some number $c$ in the interval $\left( {a,b} \right)$, such that $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$.
We know that $f\left( x \right) = {e^x}$ is a continuous function on the interval $\left[ {0,1} \right]$ and a differentiable function on the interval $\left( {0,1} \right)$. The value of $c$ can be simplified as $\ln \left( {e - 1} \right) \approx 0.54$, which lies in the interval $\left( {0,1} \right)$. Therefore, this question verifies the Mean Value Theorem.