Question

Verify Lagrange’s mean value theorem for $f(x) = \dfrac{{x + 1}}{x}$ for $x = [1,3]$.

Answer 1

Hint: To verify Lagrange’s mean value theorem, first we have to check if the given function is continuous and differentiable in the given interval. After that differentiate the function. Then we have to find the value of $c$ for which the theorem holds.

Complete step-by-step answer:
We are given the function $f(x) = \dfrac{{x + 1}}{x}$.
We have to verify Lagrange’s mean value theorem for $f(x) = \dfrac{{x + 1}}{x}$ for $x = [1,3]$.
Lagrange’s mean value theorem (LMVT) states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there is at least one point $x = c$ on this interval such that
$f(b) - f(a) = f'(c)(b - a)$
Here $a = 1,b = 3$
We have to check the function is continuous on a closed interval $[1,3]$ and differentiable on the open interval $(1,3)$.
Every rational function is continuous.
A rational function is differentiable if the denominator is not equal to zero.
A rational function is of the form $\dfrac{{p(x)}}{{q(x)}}$, where $p(x),q(x)$ are polynomials.
So the given function is continuous and differentiable.
Consider $f(x) = \dfrac{{x + 1}}{x}$
The derivative of a function of the form $y = \dfrac{{g(x)}}{{h(x)}}$ is given by $\dfrac{{dy}}{{dx}} = \dfrac{{h(x)g'(x) - g(x)h'(x)}}{{{{(h(x))}^2}}}$.
Differentiating with respect to $x$,
$f'(x) = \dfrac{{x \times \dfrac{{d(x + 1)}}{{dx}} - (x + 1)\dfrac{{dx}}{{dx}}}}{{{x^2}}}$
We know $\dfrac{{d(x + 1)}}{{dx}} = 1$ and $\dfrac{{dx}}{{dx}} = 1$
$ \Rightarrow f'(x) = \dfrac{{x \times 1 - (x + 1) \times 1}}{{{x^2}}}$
Simplifying we get,
$ \Rightarrow f'(x) = \dfrac{{x - (x + 1)}}{{{x^2}}}$
$ \Rightarrow f'(x) = \dfrac{{x - x - 1}}{{{x^2}}} = \dfrac{{ - 1}}{{{x^2}}}$
So for some $c \in (1,3)$
$f'(c) = \dfrac{{ - 1}}{{{c^2}}}$
Let us check the value of $c$ for which the theorem holds.
We have,
$f(b) - f(a) = f'(c)(b - a)$
$ \Rightarrow f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
Substituting values for $a,b$ we get,
$ \Rightarrow f'(c) = \dfrac{{f(3) - f(1)}}{{3 - 1}}$
$ \Rightarrow f'(c) = \dfrac{{f(3) - f(1)}}{2}$
Since $f(x) = \dfrac{{x + 1}}{x}$
We have,
$f(3) = \dfrac{{3 + 1}}{3} = \dfrac{4}{3}$ and $f(1) = \dfrac{{1 + 1}}{1} = 2$
Substituting we get,
$ \Rightarrow f'(c) = \dfrac{{\dfrac{4}{3} - 2}}{2}$
$ \Rightarrow f'(c) = \dfrac{{\dfrac{{4 - 6}}{3}}}{2}$
Simplifying we get,
$ \Rightarrow f'(c) = \dfrac{{ - 2}}{6} = \dfrac{{ - 1}}{3}$
But we have $f'(c) = \dfrac{{ - 1}}{{{c^2}}}$
Equating these two we get,
$ - \dfrac{1}{{{c^2}}} = - \dfrac{1}{3}$
$ \Rightarrow \dfrac{1}{{{c^2}}} = \dfrac{1}{3}$
Taking reciprocals,
${c^2} = 3$
Taking square roots on both sides we get,
$c = \pm \sqrt 3 $
Since $ - \sqrt 3 $ does not belong to $(1,3)$, we get $c = \sqrt 3 $.
That is for $\sqrt 3 \in (1,3)$, $f'(\sqrt 3 ) = \dfrac{{ - 1}}{3}$ and $\dfrac{{f(3) - f(1)}}{{3 - 1}} = - \dfrac{1}{3}$
So the theorem is verified.

Note: Lagrange’s mean value theorem (LMVT) states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there is at least one point $x = c$ on this interval such that
$f(b) - f(a) = f'(c)(b - a)$
Every rational function is continuous.
A rational function is differentiable if the denominator is not equal to zero.
The derivative of a function of the form $y = \dfrac{{g(x)}}{{h(x)}}$ is given by $\dfrac{{dy}}{{dx}} = \dfrac{{h(x)g'(x) - g(x)h'(x)}}{{{{(h(x))}^2}}}$.