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

We have $f(x) = x^2 - 3x + 2$
Since a polynomial function is everywhere continuous and differentiable.
Therefore, $f(x)$ is continuous on $-1, 2$ and differentiable on $-1, 2.$
Thus, both conditions of Lagrange's mean value theorem are satisfied.
So, there must exist at least one real number $\text{c}\in-1,2$ such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2+1}$
$\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{3}$
Now, $f(x) = x^2 - 3x + 2$
$\Rightarrow f'(x) = 2x - 3$
$\Rightarrow f(2) = 0$
$\Rightarrow f(-1) = (-1)^2 - 3(-1) + 2$
$\Rightarrow f(-1) = 6$
$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(2)-\text{f}(-1)}{3}$
$\Rightarrow2\text{x}-3=-2$
$\Rightarrow2\text{x}-1=0$
$\Rightarrow\text{x}=\frac{1}{2}$
Thus, $\text{c}=\frac{1}{2}\in(1,2)$ 
such that $\text{f}'(\text{c})=\frac{\text{f}(2)-\text{f}(-1)}{2-(-1)}$
Hence, Lagrange's mean value theorem is verified.

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