CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Verify Rolle's theorem for the following function on the indicated intervals $f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
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Answer
Here,$f(x) = (x^2- 1)(x - 2)$ on $[-1, 2]$
$f(x)$ is continuous is $[-1, 2]$ and differentiable in $(-1, 2)$ as it is a polynomial functions.
Now,
$f(-1) = (1-1)(-1-2) = 0$
$f(2) = (4-1)(2-2) = 0$
$\Rightarrow f(-1) = f(2)$
So, Rolle's theorem is applicable on $f(x)$ is $[-1, 2]$
therefore, we have to show that there exist a $\text{c}\in(-1,2)$ such that $f'(c) = 0$
Now,
$f(x) = (x^2- 1)(x - 2)$
$f'(x) = 2x(x - 2) + (x^2 - 1)$
$= 2x^2 - 4 + x^2 - 1$
$f'(x) = 3x^2 - 5$
Now,
$f'(c) = 0$
$\Rightarrow 3x^2 - 5 = 0$
$\Rightarrow\text{x}=-\sqrt{\frac{5}{3}}$ or $\text{x}=\sqrt{\frac{5}{3}}\in(-1,2)$
Thus, Rolle's theorem is verified.
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