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 function

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

in

[0, 4]

.

Accepted Answer

Given,

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

$\Rightarrow f (x) = (x^{2} - 3 x + 2) (x - 3)$

$\Rightarrow f (x) = x^{3} - 6 x^{2} + 11 x - 6 . . . . . (1)$

To check applicability of Lagrange's Mean Value Theorem for the function in $[0, 4]$ , we need to check following two conditions :

$(i)$ Continuity of $f (x)$ in $[0, 4]$ .

$(ii)$ Differentiability of $f (x)$ in $(0, 4)$

Here, $f (x) = x^{3} - 6 x^{2} + 11 x - 6$ being a polynomial function, it is continuous and differential for all real values of $x$ .

$\Rightarrow f (x)$ is continuous in the closed interval $[0, 4]$ and differentiable in open interval $(0, 4)$ .

Thus, $f (x)$ satisfied both the condition of Lagrange's Mean Value Theorem.

$∴$ There must exist at least one number $c \in (0, 4)$ such that

$f^{'} (c) = \frac{f (4) - f (0)}{4 - 0} . . . . (2)$

$f (4) = 6, f (0) = - 6$

Differentiating both sides of equation $(1)$ with respect to $x$ , we get

$f' (x) = 3 x^{2} - 12 x + 11$

Put $x = c$ , then

$f^{'} (c) = 3 c^{2} - 12 c + 11$

Hence, by $(2)$

$3 c^{2} - 12 c + 11 = \frac{6 - (- 6)}{4 - 0}$

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

By quadratic formula,

$c = \frac{12 \pm \sqrt{{(12)}^{2} - 4 \times 3 \times 8}}{12}$

$\Rightarrow c = \frac{12 \pm \sqrt{144 - 96}}{6}$

$\Rightarrow c = \frac{12 \pm 6.92}{6}$

$\Rightarrow c = 3.15, 0.84$

Both values lie in open interval $(0, 4)$

Hence, Lagrange's Mean Value theorem is verified.

Verify Lagrange's mean value theorem for the function $f (x) = (x - 1) (x - 2) (x - 3)$ in $[0, 4]$ .

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Verify Lagrange's mean value theorem for the function f(x)=x-1x-2x-3 in 0,4.

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

Verify Lagrange's mean value theorem for the function $f (x) = (x - 1) (x - 2) (x - 3)$ in $[0, 4]$ .