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CONTINUITY AND DIFFERENTIABILITY — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSCONTINUITY AND DIFFERENTIABILITY3 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.
$f(x) = 2x^2 - 3x + 1$ on $[1, 3]$
✓

Answer

Here,$f(x) = 2x^2 - 3x + 1$ on $[1, 3]$
We know that a polynomial function is continuous and differentiable.
So, $f(x)$ is continuous in $[1, 3]$ and $f(x)$ differentiable in $(1, 3).$
So, Lagrange's mean value theorem is applicable.
So, there must exist at least one real number $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(-1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{(2(3)^2-3(3)+1)-(2-3+1)}{3-1}$
$\Rightarrow4\text{c}-3=\frac{10}{2}$
$\Rightarrow4\text{c}=5+3$
$\Rightarrow4\text{c}=8$
$\Rightarrow\text{c}=2\in(1,3)$
Hence, Lagrange's mean value theorem is verified.

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