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  • Verify mean value theorem for each of the functions given f(x) = 1/4x-1

Verify mean value theorem for each of the functions given
f(x)=\frac{1}{4 x-1} \text { in }[1,4]

Answers (1)

Given: f(x)=\frac{1}{4 x-1} \text { in }[1,4]

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
$$ f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
Here,
$$ f(x)=\frac{1}{4 x-1}=(4 x-1)^{-1}
On differentiating above with respect to x, we get
\\f^{\prime}(x)=-1 \times(4 x-1)^{-1-1} \times 4 \\\Rightarrow f^{\prime}(x)=-4 \times(4 x-1)^{-2} \\\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{-4}{(4 \mathrm{x}-1)^{2}} \\\Rightarrow f^{\prime}(x) exist
Hence, f(x) is differentiable in (1,4)
We know that,
Differentiability \Rightarrow Continuity
Hence,  f(x)  is continuous in (1,4)

Thus, Mean Value Theorem is applicable to the given function

Now,

\\ f(x)=\frac{1}{4 x-1} x \in[1,4] \\ f(a)=f(1)=\frac{1}{4(1)-1}=\frac{1}{4-1}=\frac{1}{3} \\ f(b)=f(4)=\frac{1}{4(4)-1}=\frac{1}{16-1}=\frac{1}{15}

Now, let us show that there exist c ∈ (0,1) such that

\\f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f(x)=\frac{1}{4 x-1}$ \\On differentiating above with respect to $x,$ we get \\$f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ \\Put $x=c$ in above equation, we get \\$f^{\prime}(c)=\frac{-4}{(4 c-1)^{2}}$ \\By Mean Value Theorem, \\$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ \\$f^{\prime}(c)=\frac{f(4)-f(1)}{4-1}

\\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{\frac{1}{15}-\frac{1}{3}}{3} \\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{\frac{1-5}{15}}{3} \\ \Rightarrow \frac{-4}{(4 c-1)^{2}}=\frac{-4}{15 \times 3} \\ \Rightarrow(4 c-1)^{2}=45 \\ \Rightarrow 4 c-1=\sqrt{ 45} \\ \Rightarrow 4 c-1=\pm 3 \sqrt{5}

\\\Rightarrow 4 c=1 \pm 3 \sqrt{5} \\\Rightarrow c=\frac{1 \pm 3 \sqrt{5}}{4}
but c=\frac{1-3 \sqrt{5}}{4} \notin(1,4)
So, value of \mathrm{c}=\frac{1+3 \sqrt{5}}{4} \in(1,4)

Thus, Mean Value Theorem is verified.

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infoexpert22

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