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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.
$f(x) = x^3 - 2x^2 - x + 3 on [0, 1]$
✓

Answer

Here,$f(x) = x^3 - 2x^2 - x + 3$
Since $f(x)$ is polynomial function. So, $f(x)$ is continuous in $[0, 1]$ and differentiable in $(0, 1).$
Thus, both conditions of Lagrange's mean value theorem is appplicable.
Thus, there exist a point $\text{c}\in(0,1)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$
$\Rightarrow 3c^2 - 4c - 1 = 1^{-2}$
$\Rightarrow 3c^2 - 4c + 1 = 0$
$\Rightarrow 3c^2- 3c - c + 1 = 0$
$\Rightarrow 3c(c - 1) - 1(c - 1) = 0$
$\Rightarrow (3c - 1)(c - 1) = 0$
$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$
Hence, Lagrange's mean value theorem is verified.

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