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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) = X2 − 2x + 4 on [1, 5] ? - 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) = x2 − 2x + 4 on [1, 5] ?

Solution

We have,

$f\left( x \right) = x^2 - 2x + 4$

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

$f\left( x \right)$ is continuous on $\left[ 1, 5 \right]$ and differentiable on $\left( 1, 5 \right)$ .
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number
$c \in \left( 1, 5 \right)$ such that $f'\left( c \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{5 - 1} = \frac{f\left( 5 \right) - f\left( 1 \right)}{4}$
Now,
$f\left( x \right) = x^2 - 2x + 4$
$\Rightarrow f'\left( x \right) = 2x - 2$ ,
$f\left( 5 \right) = 25 - 10 + 4 = 19$ ,
$f\left( 1 \right) = 1 - 2 + 4 = 3$
∴  $f'\left( x \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{4}$

$\Rightarrow 2x - 2 = \frac{19 - 3}{4}$

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

$\Rightarrow x = \frac{6}{2} = 3$

Thus,

$c = 3 \in \left( 1, 5 \right)$ such that

$f'\left( c \right) = \frac{f\left( 5 \right) - f\left( 1 \right)}{5 - 1}$ .

Hence, Lagrange's theorem is verified.

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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) = X2 − 2x + 4 on [1, 5] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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