Question

Solve for the inverse of the dilated and translated quadratic function f(x)=7x2+5
with domain restricted to x≥0
.(1 point)
Responses

f−1(x)=7x−5−−−−−√
f inverse left parenthesis x right parenthesis equals Start Root 7 x minus 5 End Root

f−1(x)=x−57−−−√
f inverse left parenthesis x right parenthesis equals Start Root Start Fraction x minus 5 over 7 End Fraction End Root

f−1(x)=7x−−√+5
f inverse left parenthesis x right parenthesis equals Start Root 7 x End Root plus 5

f−1(x)=x+57−−−√

Answer 1

To find the inverse of the function \( f(x) = 7x^2 + 5 \) with the domain restricted to \( x \geq 0 \), we follow these steps:

1. **Rearrange the function**: Start by setting \( y = f(x) \):
\[
y = 7x^2 + 5
\]

2. **Isolate \( x \)**: First, we need to express \( x \) in terms of \( y \). Rearrange the equation to solve for \( x \):
\[
y - 5 = 7x^2
\]
\[
\frac{y - 5}{7} = x^2
\]
\[
x = \sqrt{\frac{y - 5}{7}}
\]
(We take the positive root since \( x \geq 0 \).)

3. **Write the inverse function**: Now we will express this in terms of \( x \):
\[
f^{-1}(x) = \sqrt{\frac{x - 5}{7}}
\]

Now, let’s identify which of the responses matches our derived inverse function:

- \( f^{-1}(x) = \sqrt{\frac{x - 5}{7}} \)

This matches the response:

\[
f^{-1}(x) = \frac{\sqrt{x - 5}}{\sqrt{7}}
\]

Thus, the correct option that represents the inverse function is:

\[
f^{-1}(x) = \sqrt{\frac{x - 5}{7}}
\]