Question

Find the inverse of the translated quadratic function f(x)=2x^2+8, with domain restricted to [0,∞)
a. f^-1= (√1/2x=8)
b. f^-1= (√1/2x-4)
c. f^-1= 1/2(√x-4)
d. f^-1= 1/2(√x-8)

Answer 1

To find the inverse of the function \( f(x) = 2x^2 + 8 \) with the domain restricted to \( [0, \infty) \), we follow these steps:

1. **Replace \( f(x) \) with \( y \)**:
\[
y = 2x^2 + 8
\]

2. **Solve for \( x \) in terms of \( y \)**:
\[
y - 8 = 2x^2
\]
\[
\frac{y - 8}{2} = x^2
\]
\[
x = \sqrt{\frac{y - 8}{2}} \quad (\text{since } x \geq 0)
\]

3. **Swap \( x \) and \( y \) to find the inverse function**:
\[
f^{-1}(x) = \sqrt{\frac{x - 8}{2}}
\]

Next, let's express it in a more standard form:
\[
f^{-1}(x) = \frac{1}{\sqrt{2}} \sqrt{x - 8}
\]

Now to match with the given options:

If we multiply both the numerator and denominator by \(\sqrt{2}\):
\[
f^{-1}(x) = \frac{1}{2}(\sqrt{2(x - 8)}) = \frac{1}{2}(\sqrt{2x - 16})
\]

However, since \(f^{-1}(x) = \sqrt{\frac{x - 8}{2}}\) doesn't seem to match any of the original options directly, let's analyze the given alternative options again for closeness:

- a. \( f^{-1}(x) = \sqrt{\frac{1}{2} x - 8} \)
- b. \( f^{-1}(x) = \sqrt{\frac{1}{2} x - 4} \)
- c. \( f^{-1}(x) = \frac{1}{2}(\sqrt{x} - 4) \)
- d. \( f^{-1}(x) = \frac{1}{2}(\sqrt{x} - 8) \)

None of these seem to perfectly convey our expression. However, observing **Option b**, we can evaluate its validity by substituting values:

If we check option b:
\[
f^{-1}(x) = \sqrt{\frac{1}{2} x - 4}
\]
Evaluating for a specific range, it is important and helps to determine which expression seems conceptually closer when we simplify.

Thus reviewing carefully, we determined that while none of the answers can currently substantiate as exact, option b seems to be more typical for practical aspect.

Hence, checking \(b\) yields default best match:
### **Answer:**
**b. f^-1(x) = √(1/2)x - 4** (with necessary margin adjustments for defined constants).