Question

Which of the following is the inverse of the function f(x)=6x2+12
𝑓
(
𝑥
)
=
6
𝑥
2
+
12
with a domain restricted to x≥0
𝑥
≥
0
?(1 point)
Responses

f−1(x)=x−12−−−−−√
𝑓
−
1
(
𝑥
)
=
𝑥
−
12
f inverse of x is equal to square root of x minus 12 end root

f−1(x)=x6−12−−−−−−√
𝑓
−
1
(
𝑥
)
=
𝑥
6
−
12
f inverse of x is equal to square root of x over 6 minus 12 end root

f−1(x)=x6+2−−−−−√
𝑓
−
1
(
𝑥
)
=
𝑥
6
+
2
f inverse of x is equal to square root of x over 6 plus 2 end root

f−1(x)=x6−2−−−−−√

Answer 1

To find the inverse of the function \( f(x) = 6x^2 + 12 \) with the restriction \( x \geq 0 \), we will first set \( y = f(x) \):

\[
y = 6x^2 + 12
\]

Next, we can solve for \( x \) in terms of \( y \):

1. Subtract 12 from both sides:
\[
y - 12 = 6x^2
\]

2. Divide both sides by 6:
\[
\frac{y - 12}{6} = x^2
\]

3. Take the square root of both sides. Since \( x \) is non-negative (due to the domain restriction), we take the positive root:
\[
x = \sqrt{\frac{y - 12}{6}}
\]

Now we replace \( y \) with \( x \) to express the inverse function:

\[
f^{-1}(x) = \sqrt{\frac{x - 12}{6}}
\]

Thus, the correct form of the inverse function is:

\[
f^{-1}(x) = \sqrt{\frac{x - 12}{6}}
\]

Looking at the provided options:

1. **Incorrect:** \( f^{-1}(x) = \sqrt{x - 12} \)
2. **Incorrect:** \( f^{-1}(x) = \sqrt{\frac{x}{6}} - 12 \)
3. **Incorrect:** \( f^{-1}(x) = \sqrt{\frac{x}{6}} + 2 \)
4. **Correct:** \( f^{-1}(x) = \sqrt{\frac{x - 12}{6}} \)

Since the last response corresponds to the correct inverse function, the correct answer is that:

\[
f^{-1}(x) = \sqrt{\frac{x - 12}{6}}
\]