PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S Mean F(X) = X(X −1) on [1, 2] ? - PUC Karnataka Science Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem f(x) = x(x −1) on [1, 2] ?

#### Solution

We have,

$f\left( x \right) = x\left( x - 1 \right)$ which can be rewritten as $f\left( x \right) = x^2 - x$

Since a polynomial function is everywhere continuous and differentiable.
Therefore,  $f\left( x \right)$ is continuous on $\left[ 1, 2 \right]$ and differentiable on $\left( 1, 2 \right)$
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ $c \in \left( 1, 2 \right)$ such that

$f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$

Now,

$f\left( x \right) = x^2 - x$

$\Rightarrow f'\left( x \right) = 2x - 1$,
$f\left( 2 \right) = 2$ ,
$f\left( 1 \right) = 0$
∴  $f'\left( x \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$

$\Rightarrow 2x - 1 = \frac{2 - 0}{2 - 1}$

$\Rightarrow 2x - 1 - 2 = 0$

$\Rightarrow 2x = 3$

$\Rightarrow x = \frac{3}{2}$

Thus,

$c = \frac{3}{2} \in \left( 1, 2 \right)$ such that

$f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1}$ .
Hence, Lagrange's theorem is verified.
Is there an error in this question or solution?

#### APPEARS IN

Solution Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S Mean F(X) = X(X −1) on [1, 2] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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