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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
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 - 1 on [2, 3]
✓

Answer

We have
f(x) = x2 - 1
Since a polynomial function is everywhere continuous and differentiable, f(x) is continuous on 2, 3 and differentiable on 2, 3.
Thus, both conditions of Lagrange's mean value theorem is satisfied.
So, there must exist at least one real number $\text{c}\in2,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
Now,
f(x) = x2 - 1
⇒ f'(x) = 2x,
⇒ f(3) = (3)2 - 1 = 8
⇒ f(2) = (2)2 - 1 = 3
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
$\Rightarrow2\text{x}=\frac{8-3}{1}$
$\Rightarrow\text{x}=\frac{5}{2}$
Thus,
$\text{c}=\frac{5}{2}\in(2,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$

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