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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 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.
$f(x) = x^2- 1$ on $[2, 3]$
✓

Answer

We have
$f(x) = x^2- 1$
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,
$f(x) = x^2 - 1$
$\Rightarrow f'(x) = 2x,$
$\Rightarrow f(3) = (3)^2 - 1 = 8$
$\Rightarrow f(2) = (2)^2 - 1 = 3$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(3)-\text{f}(2)}{3-2}$
$\Rightarrow2\text{x}=\frac{8-3}{1}$
$\Rightarrow\text{x}=\frac{5}{2}$
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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