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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 + 4)2 on [0, 4 - 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) = x(x + 4)2 on [0, 4] ?

#### Solution

We have,

$f\left( x \right) = x \left( x + 4 \right)^2 = x\left( x^2 + 16 + 8x \right) = x^3 + 8 x^2 + 16x$

Since

$f\left( x \right)$  is a polynomial function which is everywhere continuous and differentiable.

Therefore,

$f\left( x \right)$ is continuous on $\left[ 0, 4 \right]$ and derivable on $\left( 0, 4 \right)$

Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some

$c \in \left( 0, 4 \right)$ such that
$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0} = \frac{f\left( 4 \right) - f\left( 0 \right)}{4}$
Now,
$f\left( x \right) = x^3 + 8 x^2 + 16x$
$f'\left( x \right) = 3 x^2 + 16x + 16$,
$f\left( 4 \right) = 64 + 128 + 64 = 256$ ,
$f\left( 0 \right) = 0$
∴  $f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}$

$\Rightarrow 3 x^2 + 16x + 16 = \frac{256}{4}$

$\Rightarrow 3 x^2 + 16x - 48 = 0$

$\Rightarrow x = - \frac{4}{3}\left( 2 + \sqrt{13} \right), \frac{4}{3}\left( \sqrt{13} - 2 \right)$

Thus,

$c = \frac{- 8 + 4\sqrt{13}}{3} \in \left( 0, 4 \right)$ such that

$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}$.

Hence, Lagrange's theorem is verified.

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#### APPEARS IN

Solution for 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 F(X) = X(X + 4)2 on [0, 4 concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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