Q. Which equation is the inverse of \(y = 16x^2 + 1\)?

Answer

Swap \(x\) and \(y\): \(x=16y^{2}+1\).

Solve for \(y\):

\[ y=\pm\sqrt{\frac{x-1}{16}}=\pm\frac{1}{4}\sqrt{x-1} \]

For an inverse function, restrict the original domain to \(x\geq 0\) to get:

\[ y=\frac{1}{4}\sqrt{x-1} \]

(or \(x\leq 0\) to get \(y=-\frac{1}{4}\sqrt{x-1}\)).

Detailed Explanation

Problem

Find the inverse of the function \(y = 16x^2 + 1\).

Check one-to-one: The given function is a parabola opening upward, so it is not one-to-one on all real x (fails the horizontal line test). To have an inverse that is a function, we must restrict the domain. Common restrictions are x ≥ 0 or x ≤ 0.
Swap x and y to begin solving for the inverse:

\(x = 16y^2 + 1\)
Solve for y step by step:
1. Subtract 1 from both sides: \(x – 1 = 16y^2\)
2. Divide by 16: \(y^2 = \dfrac{x – 1}{16}\)
3. Take square roots: \(y = \pm \sqrt{\dfrac{x – 1}{16}}\)
4. Simplify the radical: \(y = \pm \dfrac{1}{4}\sqrt{x – 1}\)
Choose the correct sign according to the domain restriction:
- If the original function is restricted to x ≥ 0, then the inverse must produce nonnegative outputs, so take the positive branch:
  \(y = \dfrac{1}{4}\sqrt{x – 1}\) with domain \(x \ge 1\) (because the original range is \([1,\infty)\)).
- If the original function is restricted to x ≤ 0, then take the negative branch:
  \(y = -\dfrac{1}{4}\sqrt{x – 1}\) with domain \(x \ge 1\).
Conclusion — the inverse(s):

With a domain restriction x ≥ 0 for the original function, the inverse is
\(y = \dfrac{1}{4}\sqrt{x – 1}\).

With a domain restriction x ≤ 0 for the original function, the inverse is
\(y = -\dfrac{1}{4}\sqrt{x – 1}\).

Without restricting the original domain, the relation \(y = \pm \dfrac{1}{4}\sqrt{x – 1}\) is not a function.

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FAQs

What is the inverse relation of \(y = 16x^2 + 1\)?

Swap \(x\) and \(y\): \(x = 16y^2 + 1\). Solve: \(y = \pm\sqrt{\frac{x-1}{16}} = \pm \frac{1}{4}\sqrt{x-1}\). Domain of the relation: \(x \ge 1\).

Is the inverse a function?

Not on the full domain: the relation has two outputs (\(+\) and \(-\)) for each \(x \ge 1\). To be a function you must restrict the original domain (e.g., \(x \ge 0\) or \(x \le 0\)).

How do I get a true inverse function (steps)?

Steps: 1) Write \(y=16x^2+1\). 2) Swap: \(x=16y^2+1\). 3) Solve: \(y^2 = \frac{x-1}{16}\). 4) Take square root: \(y = \pm\frac{1}{4}\sqrt{x-1}\). 5) Restrict sign by restricting original domain.

If I restrict the domain to \(x\ge0\), what is \(f^{-1}(x)\)?

With original domain \(x \ge 0\): \(f^{-1}(x) = \frac{1}{4}\sqrt{x-1}\), domain \(x \ge 1\), range \(y \ge 0\).

What are the domain and range of the inverse relation/function?

Relation domain: \([1,\infty)\). Relation range: \((-\infty,0]\cup[0,\infty)\). For the positive-branch inverse (from \(x\ge0\)): domain \([1,\infty)\), range \([0,\infty)\). For the negative branch (from \(x\le0\)): range \((-\infty,0]\).

How can I verify the inverse is correct?

How can I verify the inverse is correct?

How does the graph of the inverse relate to the original graph?

The inverse is the reflection of \(y=16x^2+1\) across the line \(y=x\). The parabola splits into two reflected half-curves: \(y=\pm\frac{1}{4}\sqrt{x-1}\) (only \(x \ge 1\)).

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