Question

Question
Solve for the inverse of the dilated and translated quadratic function f(x)=5x2−4
with domain restricted to x≥0
.(1 point)
Responses

f−1(x)=5x24−−−√
f inverse left parenthesis x right parenthesis equals Start Root Start Fraction 5 x squared over 4 End Fraction End Root

f−1(x)=x−45−−−√
f inverse left parenthesis x right parenthesis equals Start Root Start Fraction x minus 4 over 5 End Fraction End Root

f−1(x)=5x2−4−−−−−−√
f inverse left parenthesis x right parenthesis equals Start Root 5 x squared minus 4 End Root

f−1(x)=x+45−−−√

Answer 1

To find the inverse of the function \( f(x) = 5x^2 - 4 \) with the domain restricted to \( x \geq 0 \), follow these steps:

1. **Write the function with \( y \)**:
\[
y = 5x^2 - 4
\]

2. **Solve for \( x \)** in terms of \( y \):
\[
y + 4 = 5x^2
\]
\[
\frac{y + 4}{5} = x^2
\]
\[
x = \sqrt{\frac{y + 4}{5}}
\]

Since the domain of the original function is \( x \geq 0 \), we take the positive square root.

3. **Replace \( y \) with \( x \) for the inverse function**:
\[
f^{-1}(x) = \sqrt{\frac{x + 4}{5}}
\]

Thus, the correct response is:
\[
f^{-1}(x) = \sqrt{\frac{x + 4}{5}}
\]
None of the provided answers exactly match this, so it seems there might be a mistake in the options. However, if we were to simplify, none fit as they stand, but the expected inverse is indeed:
\[
f^{-1}(x) = \sqrt{\frac{x + 4}{5}}
\]