Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

# 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 F(X) = X3 − 2x2 − X + 3 on [0, 1] ? - CBSE (Commerce) Class 12 - Mathematics

Login
Create free account

Forgot password?
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) = x3 − 2x2 − x + 3 on [0, 1] ?

#### Solution

We have,

$f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0$

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

Now,

$f\left( x \right) = x^3 - 2 x^2 - x + 3 = 0$

$\Rightarrow f'\left( x \right) = 3 x^2 - 4x - 1$ ,

$f\left( 1 \right) = 1$ ,
$f\left( 0 \right) = 3$
∴ $f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}$

$\Rightarrow 3 x^2 - 4x - 1 = \frac{1 - 3}{1}$

$\Rightarrow 3 x^2 - 4x - 1 + 2 = 0$

$\Rightarrow 3 x^2 - 4x + 1 = 0$

$\Rightarrow 3 x^2 - 3x - x + 1 = 0$

$\Rightarrow \left( 3x - 1 \right)\left( x - 1 \right) = 0$

$\Rightarrow x = \frac{1}{3}, 1$

Thus,

$c = \frac{1}{3} \in \left( 0, 1 \right)$ 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?

#### 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 F(X) = X3 − 2x2 − X + 3 on [0, 1] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
S