Resolved: "Finding inverse function of y=x+2x−1"

High School
Algebra
Algebra I
Quadratic function and equation

Willow Cooper

Answered question

2022-04-08

Finding inverse function of

y = \sqrt{x + \sqrt{2 x - 1}}

Answer & Explanation

unduncjineei5r3

Beginner2022-04-09Added 19 answers

Let $t = \sqrt{2 x - 1} \geq 0$ , then $x = \frac{t^{2} + 1}{2}$ so
$y = \sqrt{\frac{t^{2} + 2 t + 1}{2}} = \frac{t + 1}{\sqrt{2}}$
Now is easy to finish.
If I didn't make any mistake, then inverse function is
$y = x^{2} - x \sqrt{2} + 1$

glanzerjbdo

Beginner2022-04-10Added 13 answers

Step 1
$\begin{aligned} y = \sqrt{x + \sqrt{2 x - 1}} & ⟹ y^{2} = x + \sqrt{2 x - 1} \\ ⟹ y^{2} - x = \sqrt{2 x - 1} \\ ⟹ (y^{2} - x + 1)^{2} = (y^{2} - x)^{2} + 1 + 2 (y^{2} - x) \end{aligned}$
Step 2
$\begin{aligned} (y^{2} - x + 1)^{2} = 2 x - 1 + 1 + 2 \sqrt{2 x - 1} = 2 y^{2}, \end{aligned}$ and therefore $y^{2} - x + 1 = \pm \sqrt{2} y$ . So, $x = y^{2} \pm \sqrt{2} y + 1$ .

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