Question 115

If a function $$f : R \rightarrow R$$ is defined by $$f(x) = x^{2} + 1$$, then $$f^{-1}(26) =$$

Solution

$$f(x) = x^{2} + 1$$, then $$f^{-1}(26) =$$

Let f(x) = y

$$y = x^{2} + 1$$, then $$f^{-1}(26) =$$

$$x\ =\ \sqrt{\ \left(y-1\right)}$$

putting y =26

$$f^{-1}(26) =$$ = $$\sqrt{\ \left(26-1\right)}=\sqrt{\ 25}\ $$

Range = (-5,+5) Answer

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