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  • Verify the Rolle’s theorem for each of the functions f(x) = root 4-x^2

Verify the Rolle’s theorem for each of the functions
f(x)=\sqrt{4-x^{2}} \text { in }[-2,2]

Answers (1)

Given: f(x)=\sqrt{4-x^{2}}

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

Firstly, we have to show that f(x) is continuous.

Here, f(x) is continuous because f(x) has a unique value for each x ∈ [-2,2]

Condition 2:

Now, we have to show that f(x) is differentiable

\\ f(x)=\sqrt{4-x^{2}}$ \\$\Rightarrow f(x)=\left(4-x^{2}\right)^{\frac{1}{2}} \\ \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2}\left(4-\mathrm{x}^{2}\right)^{-\frac{1}{2}} \times(-2 \mathrm{x})$ [using chain rule $]$ \\$\Rightarrow f^{\prime}(x)=\frac{-x}{\sqrt{4-x^{2}}}

\\\therefore \mathrm{f}^{\prime}(\mathrm{x})$ exists for all $\mathrm{x} \in(-2,2)
So, f(x) is differentiable on (-2,2)
Hence, Condition 2 is satisfied.
Condition 3:
f(x)=\sqrt{4-x^{2}}

Now, we have to show that f(a) = f(b)

so, f(a) = f(-2)

f(-2)=\sqrt{4-(-2)^{2}}=\sqrt{4-4}=0
and f(b)=f(2)
\\ f(2)=\sqrt{4-(2)^{2}}=\sqrt{4-4}=0$ \\$\therefore f(-2)=f(2)=0
Hence, condition 3 is satisfied
Now, let us show that

\\ c \in(0,1) such that f^{\prime}(c)=0 \\f(x)=\sqrt{4-x^{2}}
On differentiating above with respect to x, we get

f^{\prime}(x)=\frac{-x}{\sqrt{4-x^{2}}}
Put  x=c  in above equation, we get,
f^{\prime}(c)=\frac{-c}{\sqrt{4-c^{2}}}
Thus, all the three conditions of Rolle's theorem is satisfied. Now we have to see that there exist c \in(-2,2) such that
\\f^{\prime}(c)=0$ \\$\Rightarrow \frac{-c}{\sqrt{4-c^{2}}}=0$ \\$\Rightarrow c=0$ \\$\because c=0 \in(-2,2)

Hence, Rolle’s theorem is verified.

Posted by

infoexpert22

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