Question

f(x) =√x and g(x)=1/√x
Verify Cauchy's mean value theorem

Answer 1

SOLUTION

TO VERIFY

Cauchy's mean value Theorem on the interval [3,5] for the functions

$\displaystyle \sf{f(x) = \sqrt{x} \: \: \: and \: \: g(x) = \frac{1}{ \sqrt{x} } }$

CONCEPT TO BE IMPLEMENTED

Let f and g be functions defined on [a,b] such that

1. f and g both are continuous in closed interval [a,b]

2. f and g are differentiable in open interval (a,b)

then there exists at least one point c ∈ (a,b) such that

$\displaystyle \sf{ \frac{f(b) - f(a)}{g(b) - g(a) } = \frac{f'(c)}{g'(c)}}$

EVALUATION

Here the given functions are

$\displaystyle \sf{f(x) = \sqrt{x} \: \: \: and \: \: g(x) = \frac{1}{ \sqrt{x} } }$

$\displaystyle \sf{f'(x) = \frac{1}{2 \sqrt{x} } \: \: \: and \: \: g'(x) = - \frac{1}{2x \sqrt{x} } }$

The given interval is [3,5]

So a = 3 & b = 5

So f and g both are continuous in closed interval [3,5]

Also f and g are differentiable in open interval (3,5)

So Cauchy's mean value Theorem is applicable

Now we have to find the point c ∈ (3,5) such that

$\displaystyle \sf{ \frac{f(b) - f(a)}{g(b) - g(a) } = \frac{f'(c)}{g'(c)}}$

$\displaystyle \sf{ \implies \: \frac{f(5) - f(3)}{g(5) - g(3) } = \frac{f'(c)}{g'(c)}}$

$\displaystyle \sf{ \implies \: \frac{ \sqrt{5} - \sqrt{3} }{ \frac{1}{ \sqrt{5} } - \frac{1}{ \sqrt{3} } } = \frac{ \frac{1}{2 \sqrt{c} } }{ - \frac{1}{2c \sqrt{c} } }}$

$\displaystyle \sf{ \implies \: \frac{ \sqrt{5} - \sqrt{3} }{ \frac{ \sqrt{3} - \sqrt{5} }{ \sqrt{15} } } = \frac{ 1 }{ - \frac{1}{c } }}$

$\displaystyle \sf{ \implies \: \frac{ \sqrt{5} - \sqrt{3} }{ \frac{ - ( \sqrt{5} - \sqrt{3}) }{ \sqrt{15} } } = \frac{ 1 }{ - \frac{1}{c } }}$

$\displaystyle \sf{ \implies \: \frac{ 1 }{ \frac{ -1 }{ \sqrt{15} } } = \frac{ 1 }{ - \frac{1}{c } }}$

$\displaystyle \sf{ \implies \: - \sqrt{15} = - c}$

$\displaystyle \sf{ \implies \:c = \sqrt{15} = 3.87 }$

Clearly c ∈ (3,5)

Hence Cauchy's mean value Theorem is verified

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Answer 2

Step-by-step explanation:

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