Question

Verify Lagrange’s Mean Value Theorem for the function f(x) = ${e^x}$ in [0,1].

Answer 1

Hint: - We have to only find the derivative of the function f(x) then put x = c, after that equate f’(c) with (f(1) – f(0)) / 1 to find the value of c.

Complete step by step solution:
As we know that for any function f(x) to satisfy LMVT in interval [a, b] f(x) should be a real valued function and it should satisfy two conditions. (i) f(x) is continuous in the interval [a, b] (ii) f(x) is also differentiable in the open interval (a, b). And there is at least one point x = c in the interval [a, b] for which f(b) – f(a) = f’(c)(b – a).
So, let us find the derivative of the function f(x) with respect to x.
$\Rightarrow$f’(x) = ${e^x}$
So, f’(c) = ${e^c}$
Now let us find the value of f(0) and f(1) because the interval is [0, 1].
So, f(0) = ${e^0} = 1$
And, f(1) = ${e^1} = e$
So, to check whether c lies in the interval [0, 1] we must equate f’(c) with f(1) – f(0).
$\Rightarrow f'\left( c \right) = \dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}}$
$\Rightarrow {e^c} = \dfrac{{e - 1}}{1}$
Cross-multiplying above equation.
$\Rightarrow {e^c} = e - 1$
Now taking log both sides of the equation.
$\Rightarrow \log \left( {{e^c}} \right) = \log \left( {e - 1} \right)$
As we know that according to the logarithmic identities $\log \left( {{e^a}} \right) = a$
$\Rightarrow$So, c = $\log \left( {e - 1} \right)$
And the value of log(e – 1) lies in the interval [0, 1]. So, there exists c such that it lies in interval [0, 1] and satisfies the LMVT condition.
Hence, LMVT is verified for the function f(x) = ${e^x}$ in [0,1].
Note: - Whenever we come up with this type of problem then first we have to find the derivative of the function f(x) and then find the value of f(a) and f(b) where a and b are the lower and upper limits of the given interval [a, b]. And then we had to find the value of c by solving the equation $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$. And if the value of c lies in the interval [a, b] then f(x) must satisfy Lagrange’s Mean Value Theorem otherwise not. This will be the easiest and efficient way to find the solution of the problem.