If f(x) = x^2 + 1,then f^{-1}(17) and f^{-1}( | vedclass

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(D) Given $f(x) = x^2 + 1$. To find the inverse,let $y = f(x) = x^2 + 1$. Solving for $x$,we get $x^2 = y - 1$,which implies $x = \pm \sqrt{y - 1}$. T

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None of these

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(D) Given $f(x) = x^2 + 1$. To find the inverse,let $y = f(x) = x^2 + 1$. Solving for $x$,we get $x^2 = y - 1$,which implies $x = \pm \sqrt{y - 1}$. Thus,$f^{-1}(y) = \pm \sqrt{y - 1}$. For $f^{-1}(17)$,we substitute $y = 17$: $f^{-1}(17) = \pm \sqrt{17 - 1} = \pm \sqrt{16} = \pm 4$. For $f^{-1}(-3)$,we substitute $y = -3$: $f^{-1}(-3) = \pm \sqrt{-3 - 1} = \pm \sqrt{-4}$. Since $\sqrt{-4}$ is not a real number,$f^{-1}(-3)$ is not defined in the set of real numbers. Therefore,the correct option is $D$ (None of these).

If $f(x) = x^2 + 1$,then $f^{-1}(17)$ and $f^{-1}(-3)$ will be

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