$\"\"$

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The function is $\"\"$ and $\"\"$ .

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The function is continuous in the interval $\"\"$ .

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$\"\"$ is continous at all points in the domain.

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$\"\"$ .

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Differentiate on each side with respect to $\"\"$ .

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$\"\"$ .

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$\"\"$

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Which is positive at all points of the domain except at the end point of the domain.

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Therefore, the function $\"\"$ is an one-to-one function and has an inverse.

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$\"\"$

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Find the inverse of the function :

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$\"\"$ .

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$\"\"$ .

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Differentiate on each side with respect to $\"\"$ .

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$\"\"$

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$\"\"$ .

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$\"\"$

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Therefore, $\"\"$ .

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Hence $\"\"$ .

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Rationalise by multiplying the denominator and numerator with $\"\"$ . $\"\"$ .

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$\"\"$

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The function $\"\"$ is an one-to-one function and has an inverse.

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$\"\"$ .