Name: Example of Mean Value Theorem
Uploaded: 2019-07-19
Duration: 2 min 17 s
Description: You've already learned about the Mean Value Theorem. Let’s explore more about the theorem with some examples in this video. Watch and learn!

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:

$f (x) = (x - 1) (x - 2) (x - 3)$ on $[0, 4]$

Accepted Answer

We have,

$f (x) = (x - 1) (x - 2) (x - 3)$

which can be rewritten as $f (x) = (x^{2} - 3 x + 2) (x - 3) = x^{3} - 6 x^{2} + 11 x - 6$

Since a polynomial function is everywhere continuous and differentiable.

Therefore, $f (x)$ is continuous on $[0, 4]$ and differentiable on $(0, 4)$ .

Thus, both conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number $c \in (0, 4)$ such that

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

Now, $f (x) = x^{3} - 6 x^{2} + 11 x - 6$

$\Rightarrow f' (x) = 3 x^{2} - 12 x + 11$ , $f (0) = - 6$ ,

$f (4) = 64 - 96 + 44 - 6 = 6$

$∴ f' (c) = \frac{f (4) - f (0)}{4 - 0}$

$\Rightarrow 3 c^{2} - 12 c + 11 = \frac{6 + 6}{4}$

$\Rightarrow 3 c^{2} - 12 c + 8 = 0$

By quadratic formula,

$c = \frac{12 \pm \sqrt{144 - 4 \times 3 \times 8}}{6}$

$\Rightarrow c = \frac{12 \pm \sqrt{48}}{6}$

$\Rightarrow c = \frac{12 \pm 4 \sqrt{3}}{6}$

$\Rightarrow c = 2 - \frac{2}{\sqrt{3}}, 2 + \frac{2}{\sqrt{3}}$

Thus, $c = 2 \pm \frac{2}{\sqrt{3}} \in (0, 4)$ such that $f' (c) = \frac{f (4) - f (0)}{4 - 0}$

Hence, Lagrange's Mean Value theorem is verified.

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:

$f (x) = (x - 1) (x - 2) (x - 3)$ on $[0, 4]$

Important Questions on Mean Value Theorems

Learn and Prepare for any exam you want !

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−1)(x−2)(x−3) on [0, 4]

Important Questions on Mean Value Theorems

Learn and Prepare for any exam you want !

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:

$f (x) = (x - 1) (x - 2) (x - 3)$ on $[0, 4]$