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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) = (x - 1)(x - 2)(x - 3) on [0, 4]
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Answer

f(x) = (x - 1)(x - 2)(x - 3) on [0, 4]
We know that, polynomial function is continuous and differentiable everywhere. So, f(x) is continuous in [0, 4] and differentiable in (0, 4). So Lagrange's mean value theorem is applicable. Thus, there exist a point $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow(\text{c}-1)(\text{c}-2)(\text{c}-3)+(\text{c}-1)(\text{c}-3)\\=\frac{(3)(2)(1)-(-1)(-2)(-3)}{4-1}$
$\Rightarrow\text{c}^2-3\text{c}+2+\text{c}^2+5\text{c}+6+\text{c}^2-4\text{c}+3=\frac{6+6}{4}$
$\Rightarrow3\text{c}^2-12\text{c}+11=3$
$\Rightarrow3\text{c}^2-12\text{c}+8=0$
$\Rightarrow\text{c}=\frac{-(-12)\pm\sqrt{144-4\times3\times8}}{6}$
$\Rightarrow\text{c}=\frac{12\pm\sqrt{48}}{6}$
$\Rightarrow\text{c}=2\pm\frac{2\sqrt3}{3}\in(0,4)$
$\Rightarrow\text{c}=2\pm\frac{2}{\sqrt3}\in(0,4)$
Hence, Lagrange's mean value theorem is verified.

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