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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat 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.
$f(x) = x(x + 4)^2$​​​​​​​ on $[0,4]$
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Answer

Here,$f(x) = x(x + 4)^2$
$\Rightarrow f(x) = x(x^2+ 16 + 8x)$
$\Rightarrow f(x) = x^3 + 8x^2+ 16x$
Since f(x) is a polynomial function which is everywhere continuous and differentiable.
Therefore, f(x) is continuous on [0, 4] and derivable on (0, 4)
Thus, both the conditions of Lagrange's theorem is satisfied.
Consequently, there exists some $\text{c}\in(0,4)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}=\frac{\text{f}(4)-\text{f}(0)}{4}$
Now,$ f(x) = x^3 + 8x^2+ 16x$
$\Rightarrow f(x) = 3x^2+ 16x + 16,$
$\Rightarrow f(4) = 64+ 128 + 64 = 256,$
$\Rightarrow f(0) = 0$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
$\Rightarrow3\text{x}^2+16\text{x}+16=\frac{256}{4}$
$\Rightarrow3\text{x}^2+16\text{x}-48=0$
$\Rightarrow\text{x}=-\frac{4}{3}\big(2+\sqrt{13}\big),\frac{4}{3}\big(\sqrt{13}-2\big)$
Thus, $\text{c}=\frac{-8+4\sqrt{13}}{3}\in(0,4)$ such that $\text{f}'(\text{c})=\frac{\text{f}(4)-\text{f}(0)}{4-0}$
Hence, Lagrange's mean value theorem is verified.

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