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 Mean F(X) = X2 + X − 1 on [0, 4] ? - CBSE (Science) 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) = x2 + x − 1 on [0, 4] ?

#### Solution

We have,

$f\left( x \right) = x^2 + x - 1$

Since polynomial function is everywhere continuous and differentiable.
Therefore,

$f\left( x \right)$ is continuous on $\left[ 0, 4 \right]$ and differentiable 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,
$f\left( x \right) = x^2 + x - 1$
$f'\left( x \right) = 2x + 1$,
$f\left( 4 \right) = 19$,
$f\left( 0 \right) = - 1$
∴ $f'\left( x \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}$

$\Rightarrow 2x + 1 = \frac{20}{4}$

$\Rightarrow 2x + 1 = 5$

$\Rightarrow 2x = 4$

$\Rightarrow x = 2$

Thus,

$c = 2 \in \left( 0, 4 \right)$ such that
$f'\left( c \right) = \frac{f\left( 4 \right) - f\left( 0 \right)}{4 - 0}$.
Hence, Lagrange's theorem is verified.
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 Mean F(X) = X2 + X − 1 on [0, 4] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
S