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# Solution for Verify the Hypothesis and Conclusion of Lagrange'S Man Value Theorem for the Function F(X) = 1 4 X − 1 , 1≤ X ≤ 4 ? - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Verify the  hypothesis and conclusion of Lagrange's man value theorem for the function
f(x) = $\frac{1}{4x - 1},$ 1≤ x ≤ 4 ?

#### Solution

The given function is $f\left( x \right) = \frac{1}{4x - 1}$.

Since for each $x \in \left[ 1, 4 \right]$ , vthe function attains a unique definite value, $f\left( x \right)$  is continuous on $\left[ 1, 4 \right]$ .

Also,

$f'\left( x \right) = \frac{- 4}{\left( 4x - 1 \right)^2}$ exists for all $x \in \left[ 1, 4 \right]$
Thus, both the conditions of Lagrange's mean value theorem are satisfied.
Consequently, there exists some
$c \in \left( 1, 4 \right)$ such that
$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1} = \frac{f\left( 4 \right) - f\left( 1 \right)}{3}$
Now,
$f\left( x \right) = \frac{1}{4x - 1}$
$\Rightarrow$ $f'\left( x \right) = \frac{- 4}{\left( 4x - 1 \right)^2}$ ,
$f\left( 4 \right) = \frac{1}{15}, f\left( 1 \right) = \frac{1}{3}$
$\therefore$ $f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}$

$\Rightarrow f'\left( x \right) = \frac{\frac{1}{15} - \frac{1}{3}}{4 - 1} = \frac{- 4}{45}$

$\Rightarrow \frac{- 4}{\left( 4x - 1 \right)^2} = \frac{- 4}{45}$

$\Rightarrow \left( 4x - 1 \right)^2 = 45$

$\Rightarrow 16 x^2 - 8x - 44 = 0$

$\Rightarrow 4 x^2 - 2x - 11 = 0$

$\Rightarrow x = \frac{1}{4}\left( 1 \pm 3\sqrt{5} \right)$

Thus,

$c = \frac{1}{4}\left( 1 + 3\sqrt{5} \right) \in \left( 1, 4 \right)$ such that

$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 1 \right)}{4 - 1}$.

Hence, Lagrange's theorem is verified.

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#### APPEARS IN

Solution Verify the Hypothesis and Conclusion of Lagrange'S Man Value Theorem for the Function F(X) = 1 4 X − 1 , 1≤ X ≤ 4 ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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