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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 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 - x2 on [0, 1]
✓

Answer

We have,

f(x) = 2x - x2 

Since a polynomial function is everywhere continuous and differentiable.

Therefore, f(x) is continuous on 0, 1 and differentiable on 0,1

Thus, both conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number $\text{c}\in0,1$ such that

$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}=\frac{\text{f}(1)-\text{f}(0)}{1}$

Now,

f(x) = 2x - x2 

⇒ f'(x) = 2 - 2x,

⇒ f(1) = 2 - 1

⇒ f(1) = 1,

⇒ f(0) = 0

$\therefore\ \text{f}'(\text{x})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$

$\Rightarrow2-2\text{x}=\frac{1-0}{1}$

$\Rightarrow-2\text{x}=1-2$

$\Rightarrow\text{x}=\frac{1}{2}$

Thus, $\text{c}=\frac{1}{2}\in(1,0)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$

Hence, Lagrange's mean value theorem is verified.

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