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Mean Value Theorems — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMean Value Theorems5 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 $[1, 3]$
✓

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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