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If f:ℝ→ℝ be d...

Question

If $f : R \to R$ be defined as $f (x) = - x^{2} + 4,$ then

A

f^{- 1} (x) = - \sqrt{4 - x}

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B

f^{- 1} (x) = \frac{4 x^{2} - 1}{x^{2}}

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C

f^{- 1} (x)

does not exist

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D

f^{- 1} (x) = \sqrt{4 - x}

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Solution

The correct option is C $f^{- 1} (x)$ does not exist

This function does not pass the horizontal test, which means it is not a one to one function. Therefore $f^{- 1} (x)$ doest not exist.

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