Question

Verify the hypothesis and conclusion of Lagrange’s mean value theorem for the function $f\left( x \right) = \dfrac{1}{{4x - 1}},\, 1 \leqslant x \leqslant 4$.

Answer 1

Lagrange’s mean value theorem is applicable if and only if the function is continuous and differentiable in the defined interval. For this first check the continuity and differentiability of the given differentiable function. The conditions for the hypothesis will be then satisfied. Next we have to verify the hypothesis as per its statement.

Complete step by step answer:
We have given $f\left( x \right) = \dfrac{1}{{4x - 1}}$ in the interval$\left[ {1,4} \right]$
We will check for the continuity and differentiability of the function in order to verify the Lagrange’s mean value theorem. $f\left( x \right)$iIs continuous in $\left[ {1,4} \right]$as the only point of discontinuity of $f\left( x \right)$ is at $\dfrac{1}{4}$which does not belong in the interval $\left[ {1,4} \right]$. Hence $f\left( x \right)$ is continuous in $\left[ {1,4} \right]$. Now the differentiation of the function is
$f'\left( x \right) = \dfrac{{ - 4}}{{{{\left( {4x - 1} \right)}^2}}}$, which exists in the interval $\left( {1,4} \right)$
So the function $f\left( x \right)$ is differentiable at $\left( {1,4} \right)$. Hence the conditions of the Lagrange’s mean value theorem are satisfied.

Now, according to the statement of the Lagrange’s mean value theorem for a continuous and differentiable function $f\left( x \right)$ in the interval $\left( {a,b} \right)$ there exist a real number $c \in \left( {a,b} \right)$ such that
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
Here we have $c \in \left( {1,4} \right)$ so, as per Lagrange’s mean value theorem we have,
$f'\left( c \right) = \dfrac{{f\left( 4 \right) - f\left( 1 \right)}}{{4 - 1}}$
Now from the function $f\left( x \right)$we have

\Rightarrow f\left( 4 \right)= \dfrac{1}{{16 - 1}} \\\ \Rightarrow f\left( 4 \right)= \dfrac{1}{{15}} \\\ $$ Also, $$f\left( 1 \right) = \dfrac{1}{{4 \times 1 - 1}} \\\ \Rightarrow f\left( 1 \right)= \dfrac{1}{{4 - 1}} \\\ \Rightarrow f\left( 1 \right)= \dfrac{1}{3} \\\ $$ Substituting the values we get $$f'\left( c \right) = \dfrac{{\dfrac{1}{{15}} - \dfrac{1}{3}}}{{4 - 1}}$$ Now as $$f'\left( x \right) = \dfrac{{ - 4}}{{{{\left( {4x - 1} \right)}^2}}}$$ so, for $$c$$ $$f'\left( c \right) = \dfrac{{ - 4}}{{{{\left( {4c - 1} \right)}^2}}}$$ On substituting for$$c$$in $$f'\left( c \right)$$ and simplifying we get, $$\dfrac{{ - 4}}{{{{\left( {4c - 1} \right)}^2}}} = \dfrac{{\dfrac{{3 - 15}}{{45}}}}{3} \\\ \Rightarrow \dfrac{{ - 4}}{{{{\left( {4c - 1} \right)}^2}}} = \dfrac{{\left( {\dfrac{{ - 12}}{{45}}} \right)}}{3} \\\ \Rightarrow \dfrac{{ - 4}}{{{{\left( {4c - 1} \right)}^2}}} = \dfrac{{ - 4}}{{15}} \\\ $$ On solving for $$c$$ $$\Rightarrow \dfrac{{ - 4}}{{{{\left( {4c - 1} \right)}^2}}} = \dfrac{{ - 4}}{{45}} \\\ \Rightarrow \dfrac{1}{{{{\left( {4c - 1} \right)}^2}}} = \dfrac{1}{{45}} \\\ \Rightarrow {\left( {4c - 1} \right)^2} = 45 \\\ $$ Taking square roots both sides we get, $$\Rightarrow \sqrt {{{\left( {4c - 1} \right)}^2}} = \sqrt {45} \\\ \Rightarrow \left( {4c - 1} \right) = \pm 3\sqrt 5 \\\ $$ Simplifying further we get, $$\Rightarrow 4c = \pm 3\sqrt 5 + 1 \\\ \Rightarrow c = \dfrac{{ \pm 3\sqrt 5 + 1}}{4} \\\ $$ Neglecting the negative value we find $$\therefore c = \dfrac{{3\sqrt 5 + 1}}{4} \in \left( {1,4} \right)$$ **Hence, the hypothesis and conclusion of Lagrange’s mean value theorem for the function $$f\left( x \right) = \dfrac{1}{{4x - 1}}, 1 \leqslant x \leqslant 4$$ is verified.** **Note:** Rolle’s Theorem is another theorem for the evaluation of C in mean theorem.The Langrage’s mean value theorem is only applicable when the given function is continuous in the given interval and also only when it is differentiable in the same open interval. A continuous function may be or may not be differentiable but a continuous function is always differentiable.