Question

Verify the applicability of LMV theorem for $f\left( x \right)$=$\dfrac{1}{{1 + x}}$ in $\left[ {1,0} \right]$

Answer 1

This theorem contains 2 steps which are discussed as: If $f$ is a real valued function on $\left[ {a,b} \right]$ such that
(i) $f$ is continuous in $\left[ {a,b} \right]$
(ii) $f$ is differentiable i.e. derivative exists in $\left[ {a,b} \right]$
Then $\exists$ at least one c $\in$ $\left( {a,b} \right)$ such that $f'\left( c \right)$= $\dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$

Complete step-by-step answer:
Here we have to verify the applicability of LMV theorem for $f\left( x \right)$=$\dfrac{1}{{1 + x}}$ in $\left[ {1,0} \right]$
LMV theorem steps are
Step 1: Since, it is polynomial $f\left( x \right)$ of the type ${a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} \ldots \ldots \ldots ...$ therefore $f\left( x \right)$ is continuous in $\left[ {1,0} \right]$ Because polynomial is continuous
Step 2: Take derivative of $f\left( x \right)$
$\Rightarrow$ $f'\left( x \right) = \; - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}} \in \left[ {1,0} \right]$
Using identity $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$
Since $f'\left( x \right)$ exists in $\left[ {1,0} \right]$ , therefore $f\left( x \right)$ is differentiable in $\left[ {1,0} \right]$
Since both conditions are satisfied.
Therefore, there exists at least one $c \in \left[ {1,0} \right]$ such that
$\Rightarrow$ $\dfrac{{f\left( 0 \right) - f\left( 1 \right)}}{{0 - 1}} = f'\left( c \right)$
Using formula $f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$ where $\left[ {a,b} \right]$ are $\left[ {1,0} \right]$
$\Rightarrow$ $\dfrac{{\dfrac{1}{{1 + 0}} - \dfrac{1}{{1 + 1}}}}{{ - 1}} = f'\left( c \right)$
$\because f(x) = \dfrac{1}{{1 + x}}$So , We get $f\left( 0 \right) = \dfrac{1}{{1 + 0}}\& f\left( 1 \right) = \dfrac{1}{{1 + 1}}$
$\Rightarrow$ $\dfrac{{\dfrac{1}{1} - \dfrac{1}{2}}}{{ - 1}} = f'\left( c \right)$
Taking LCM on numerator $\dfrac{{\dfrac{{2 - 1}}{2}}}{{ - 1}} = f'\left( c \right)$
On rearranging, we get $\dfrac{1}{2} \times - 1 = f'\left( c \right)$
Since $f'\left( x \right) = - \dfrac{1}{{{{\left( {1 + x} \right)}^2}}}$
Therefore $f'\left( c \right) = - \dfrac{1}{{{{\left( {1 + c} \right)}^2}}}$
So We get, $- \dfrac{1}{2} = - \dfrac{1}{{{{\left( {1 + c} \right)}^2}}}$
Cancelling minus from both sides
$\Rightarrow$ $\dfrac{1}{2} = \dfrac{1}{{{{\left( {1 + c} \right)}^2}}}$
Taking cross multiplication so the values become ${\left( {1 + c} \right)^2} = 2$
Taking square root on both sides
$\Rightarrow$ $1 + c = \sqrt 2$
$\Rightarrow$ $c = \sqrt 2 - 1$
$\Rightarrow$ $c = 1.414 - 1$
$\Rightarrow$ $c = 0.414 \in \left[ {1,0} \right]$

Hence LMV theorem is verified

Note: All the polynomials are continuous. All trigonometric functions like $\sin A,\cos A$ are continuous. If the derivative of $f\left( x \right) \in \left[ {a,b} \right]$ only then $f\left( x \right)$ is differentiable. Continuity should be checked on the basis of limits. If they are holding means the function is continuous as well as differentiable, it means the LMV theorem is satisfied and we can further check its verification.