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- 5x2 - 3x on [1, 3]

Answer

We have,

f(x) = x- 5x2 - 3x

Since, polynomial function is everywhere continuous and differentiable.

Therefore, f(x) is continuous on 1, 3 and differentiable on 1, 3

Thus, both the conditions of Lagrange's theorem is satisfied.

Concequently, there exist some $\text{c}\in1,3$ such that

$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$

Now, f(x) = x- 5x2 - 3x

f'(x) = 3x2 - 10x - 3

⇒ f(3) = -27

⇒ f(1) = -7

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$

$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$

$\Rightarrow3\text{x}^2-10\text{x}+7=0$

$\Rightarrow\text{x}=1,\frac{7}{3}$

Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$

Hence, Lagrange's theorem is verified.

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