For the given function $f(x)$, find $f^{-1}(x)$. State the domain of $f^{-1}(x)$.
-$f(x)=\sqrt{4 x+5}$
A) $f^{-1}(x)=\frac{2 x-5}{4}, x \geq 0$

B) $f^{-1}(x)=\frac{x^{2}}{4}-5, x \geq 0$

C) $\mathrm{f}^{-1}(\mathrm{x})=\frac{(\mathrm{x}-5)^{2}}{4}, \mathrm{x} \geq 0$

D) $f^{-1}(x)=\frac{x^{2}-5}{4}, x \geq 0$

Question

For the given function $f(x)$, find $f^{-1}(x)$. State the domain of $f^{-1}(x)$.
-$f(x)=\sqrt{4 x+5}$
A) $f^{-1}(x)=\frac{2 x-5}{4}, x \geq 0$

B) $f^{-1}(x)=\frac{x^{2}}{4}-5, x \geq 0$

C) $\mathrm{f}^{-1}(\mathrm{x})=\frac{(\mathrm{x}-5)^{2}}{4}, \mathrm{x} \geq 0$

D) $f^{-1}(x)=\frac{x^{2}-5}{4}, x \geq 0$

Quizplus · Accepted Answer

To find $f^{-1}(x)$, we set $y = \sqrt{4x+5}$, then solve for $x$ in terms of $y$. Squaring both sides gives $y^2 = 4x + 5$, leading to $x = \frac{y^2 - 5}{4}$. Thus, $f^{-1}(x) = \frac{x^2 - 5}{4}$, valid for $x \geq 0$ because the original function $f(x) = \sqrt{4x + 5}$ produces non-negative outputs for its domain.

For the Given Function f(x)f(x)f(x) , Find f−1(x)f^{-1}(x)f−1(x) State the Domain Of

For the Given Function $f(x)$ , Find $f^{-1}(x)$ State the Domain Of