Question

solve for the inverse of the dilated and translated quadratic function $f(x) = 7x^2 + 5$ with domain restricted to $x \geq 0$. (1 point) \
ewline \
ewline $\circ$ $f^{-1}(x) = \sqrt{\frac{x + 5}{7}}$ \
ewline \
ewline $\circ$ $f^{-1}(x) = \sqrt{7x - 5}$ \
ewline \
ewline $\circ$ $f^{-1}(x) = \sqrt{\frac{x - 5}{7}}$ \
ewline \
ewline $\circ$ $f^{-1}(x) = \sqrt{7x} + 5$

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( f(x) = 7x^2 + 5 \). Replace \( f(x) \) with \( y \), so we have \( y = 7x^2 + 5 \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = 7y^2 + 5 \).

Step3: Solve for \( y \)

First, subtract 5 from both sides of the equation: \( x - 5 = 7y^2 \). Then, divide both sides by 7: \( \frac{x - 5}{7} = y^2 \). Since the original domain is \( x \geq 0 \), the range of \( f(x) \) will be the domain of \( f^{-1}(x) \), and we take the positive square root (because \( x \geq 0 \) in the original function, the inverse should also be a function, so we take the positive root). So \( y = \sqrt{\frac{x - 5}{7}} \), which is \( f^{-1}(x) = \sqrt{\frac{x - 5}{7}} \).

Answer:

\( f^{-1}(x)=\sqrt{\frac{x - 5}{7}} \) (corresponding to the option \( f^{-1}(x)=\sqrt{\frac{x - 5}{7}} \))