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' F(X) = X2 − 3x + 2 on [−1, 2] ? - PUC Karnataka Science Class 12 - Mathematics

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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) = x2 − 3x + 2 on [−1, 2] ?

#### Solution

We have,

$f\left( x \right) = x^2 - 3x + 2$

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 $f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 + 1} = \frac{f\left( 2 \right) - f\left( - 1 \right)}{3}$
Now,
$f\left( x \right) = x^2 - 3x + 2$
$\Rightarrow f'\left( x \right) = 2x - 3$ ,
$f\left( 2 \right) = 0$ ,
$f\left( - 1 \right) = \left( - 1 \right)^2 - 3\left( - 1 \right) + 2 = 6$
∴  $f'\left( x \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{3}$

$\Rightarrow 2x - 3 = - 2$

$\Rightarrow 2x - 1 = 0$

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

Thus,

$c = \frac{1}{2} \in \left( - 1, 2 \right)$ such that $f'\left( c \right) = \frac{f\left( 2 \right) - f\left( - 1 \right)}{2 - \left( - 1 \right)}$ .

Hence, Lagrange's theorem is verified.

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#### 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' F(X) = X2 − 3x + 2 on [−1, 2] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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