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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. $f(x) = x^{3 }- 5x^2 - 3x$ on $[1, 3]$
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Answer

We have$, f(x) = x^{3 }- 5x^2 - 3x$
Since, polynomial function is everywhere continuous and differentiable.
Therefore$, f(x)$ is continuous on $1, 3$ and differentiable on $1, 3$
Thus, both the conditions of Lagrange's theorem is satisfied.
Concequently, there exist some $\text{c}\in1,3$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}=\frac{\text{f}(3)-\text{f}(1)}{2}$
Now$, f(x) = x^{3 }- 5x^2 - 3x$
$f'(x) = 3x^2 - 10x - 3$
$\Rightarrow f(3) = -27$
$\Rightarrow f(1) = -7$
$\therefore\ \text{f}\ '(\text{x})=\frac{\text{f}(3)-\text{f}(1)}{2}$
$\Rightarrow3\text{x}^2-10\text{x}-3=\frac{-20}{2}$
$\Rightarrow3\text{x}^2-10\text{x}+7=0$
$\Rightarrow\text{x}=1,\frac{7}{3}$
Thus, $\text{c}=\frac{7}{3}\in(1,3)$ such that $\text{f}\ '(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
Hence, Lagrange's theorem is verified.

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