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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat 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) = x2 - 3x + 2 on [-1, 2]
✓

Answer

We have

f(x) = x2 - 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) = x2 - 3x + 2

⇒ f'(x) = 2x - 3

⇒ f(2) = 0

⇒ f(-1) = (-1)2 - 3(-1) + 2

⇒ 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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