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  • Verify mean value theorem for each of the functions given f(x) = root 25-x^2 in [1,5]

Verify mean value theorem for each of the functions given
f(x)=\sqrt{25-x^2} in [1,5]

Answers (1)

Given:  f(x)=\sqrt{25-x^2}

Now, we have to show that f(x) verify the Mean Value Theorem

First of all, Conditions of Mean Value theorem are:

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

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

If both the conditions are satisfied, then there exist some ‘c’ in (a,b) such that

f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}

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 ∈ [1,5]

Condition 2:

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

\\ f(x)=\sqrt{25-x^{2}} \\ \Rightarrow f(x)=\left(25-x^{2}\right)^{\frac{1}{2}} \\ f^{\prime}(x)=\frac{1}{2}\left(25-x^{2}\right)^{-\frac{1}{2}} \times(-2 x) \\ \Rightarrow f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}}

∴ f’(x) exists for all x ∈ (1,5)

So, f(x) is differentiable on (1,5)

Hence, Condition 2 is satisfied.

Thus, mean value theorem is applicable to given function.

Now,

f(x)=\sqrt{25-x^{2}}
Now, we will find  f(a) and f(b)
so, f(a)=f(1)
f(1)=\sqrt{25-(1)^{2}}=\sqrt{25-1}=\sqrt{24}
and f(b)=f(5)
f(5)=\sqrt{25-(5)^{2}}=\sqrt{25-25}=0
Now, let us show that c \in(1,5) such that
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}

f(x)=\sqrt{25-x^{2}}
On differentiating above with respect to x, we get
f^{\prime}(x)=\frac{-x}{\sqrt{25-x^{2}}}
Put  x=c  in above equation, we get
f^{\prime}(c)=\frac{-c}{\sqrt{25-c^{2}}}
By Mean Value theorem,
\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$\Rightarrow \frac{-c}{\sqrt{25-c^{2}}}=\frac{0-\sqrt{24}}{5-1}$ \\$\Rightarrow \frac{c}{\sqrt{25-c^{2}}}=\frac{\sqrt{24}}{4}$ \\$\Rightarrow 4 c=\sqrt{24} \times \sqrt{25-c^{2}}

\begin{aligned} &\text { Squaring both sides, we get }\\ &\Rightarrow 16 c^{2}=24 \times\left(25-c^{2}\right)\\ &\Rightarrow 16 c^{2}=600-24 c^{2}\\ &\Rightarrow 24 \mathrm{c}^{2}+16 \mathrm{c}^{2}=600\\ &\Rightarrow 40 c^{2}=600\\ &\Rightarrow c^{-}=15\\ &\Rightarrow c=\sqrt{1} 5 \in(,5) \end{aligned}

Hence, Mean Value Theorem is verified.

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infoexpert22

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