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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) = 2x − X2 on [0, 1] ? - CBSE (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) = 2x − x2 on [0, 1] ?

#### Solution

We have,

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

Since a polynomial function is everywhere continuous and differentiable.
Therefore,

$f\left( x \right)$ is continuous on $\left[ 0, 1 \right]$ and differentiable on $\left( 0, 1 \right)$ .

Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number ​ $c \in \left( 0, 1 \right)$ such that

$f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0} = \frac{f\left( 1 \right) - f\left( 0 \right)}{1}$

Now,

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

$\Rightarrow f'\left( x \right) = 2 - 2x$,

$f\left( 1 \right) = 2 - 1 = 1$,

$f\left( 0 \right) = 0$

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

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

$\Rightarrow - 2x = 1 - 2$

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

Thus,

$c = \frac{1}{2} \in \left( 0, 1 \right)$ such that

$f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}$ .

Hence, Lagrange's theorem is verified.

Is there an error in this question or solution?

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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) = 2x − X2 on [0, 1] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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