Inverse function

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Randyyy

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Hey, I am asked to find the inverse of [MATH]y = x\sqrt{2+x^2} [/MATH] if such exist.
x = [MATH]y\sqrt{2+y^2} [/MATH][MATH]y = \frac{x}{\sqrt{2+x^2}}[/MATH] , I know that if x is positive, so is y because [MATH]\sqrt{2+x^2}[/MATH] is always [MATH]\geq[/MATH] 0 no matter x, and it is multiplied by x so the sign in front of y is dependent on the sign in front of x. But how do I go on from here?
 
Oh yea, you are right, that is an error on my part. I flipped my x and y variables and forgot that I no longer had an x under the square root

[MATH]x = y \sqrt{2+y^2}[/MATH][MATH]x^2=y^2(2+y^2)[/MATH][MATH]x^2=2y^2+y^4[/MATH]But how do I continue if I want to isolate y?
 
Randyyy said:
Hey, I am asked to find the inverse of [MATH]y = x\sqrt{2+x^2} [/MATH] if such exist.
x = [MATH]y\sqrt{2+y^2} [/MATH][MATH]y = \frac{x}{\sqrt{2+x^2}}[/MATH] , I know that if x is positive, so is y because [MATH]\sqrt{2+x^2}[/MATH] is always [MATH]\geq[/MATH] 0 no matter x, and it is multiplied by x so the sign in front of y is dependent on the sign in front of x. But how do I go on from here?
Click to expand...
Look at the plot. The function is one-to-one, so the inverse exists. See HERE
 
Randyyy said:
Oh yea, you are right, that is an error on my part. I flipped my x and y variables and forgot that I no longer had an x under the square root

[MATH]x = y \sqrt{2+y^2}[/MATH][MATH]x^2=y^2(2+y^2)[/MATH][MATH]x^2=2y^2+y^4[/MATH]But how do I continue if I want to isolate y?
Click to expand...
Substitute:

u = y2 then you get:

[MATH]x^2=2y^2+y^4 \ \to \ x^2 = 2u + u^2[/MATH]
Now solve the ensuing quadratic equation in 'u' and consequently solve for 'y'.
 
[MATH]u^2+2u-x^2 = 0[/MATH][MATH]u= 1 \pm \sqrt{1^2+x^2}[/MATH][MATH]y=\pm \sqrt{1 \pm \sqrt{1^2+x^2}}[/MATH]So I know that y is either positive or negative because of f(x). So the first [MATH]\pm[/MATH] gives me one option, not sure how I deal with the [MATH]\pm[/MATH] under the square roots. can both be valid depending on if my inverse is >0 or 0<?
 
Randyyy said:
[MATH]u^2+2u-x^2 = 0[/MATH][MATH]u= 1 \pm \sqrt{1^2+x^2}[/MATH][MATH]y=\pm \sqrt{1 \pm \sqrt{1^2+x^2}}[/MATH]So I know that y is either positive or negative because of f(x). So the first [MATH]\pm[/MATH] gives me one option, not sure how I deal with the [MATH]\pm[/MATH] under the square roots. can both be valid depending on if my inverse is >0 or 0<?
Click to expand...
Check "sign" of the expressions that you have derived.. Can 'y' or 'x' be "complex" numbers?
 
Last edited by a moderator:
I overlooked that fact. You are right.
[MATH] y = \pm \sqrt{1+\sqrt{1^2+x^2}}[/MATH] is really my only solution because [MATH]\sqrt{1- \sqrt{1+x^2}} < 0[/MATH]So how do I pick whether or not I have [MATH] y = \sqrt{1+\sqrt{1^2+x^2}}[/MATH] or if
[MATH] y = - \sqrt{1+\sqrt{1^2+x^2}}[/MATH] or is it so that both solutions can be valid but only one at a time because x and y share the same sign.
 
Randyyy said:
I overlooked that fact. You are right.
[MATH] y = \pm \sqrt{1+\sqrt{1^2+x^2}}[/MATH] is really my only solution because [MATH]\sqrt{1- \sqrt{1+x^2}} < 0[/MATH]So how do I pick whether or not I have [MATH] y = \sqrt{1+\sqrt{1^2+x^2}}[/MATH] or if
[MATH] y = - \sqrt{1+\sqrt{1^2+x^2}}[/MATH] or is it so that both solutions can be valid but only one at a time because x and y share the same sign.
Click to expand...
You should have gotten:

u=-1±√(12+x2) ....... you missed the "-" at the front. Please fix it.

As you have presented the problem, both the answers are acceptable.
 
[MATH]u=-1 \pm \sqrt{(-1)^2+x^2}[/MATH][MATH]y=\pm \sqrt{-1 + \sqrt{(-1)^2+x^2}}[/MATH]I am assuming this is the error you wanted me to correct!
 
[MATH] [QUOTE="Randyyy, post: 510779, member: 80656"] [MATH]u^2+2u-x^2 = 0[/MATH][MATH]u= 1 \pm \sqrt{1^2+x^2}[/MATH][MATH]y=\pm \sqrt{1 \pm \sqrt{1^2+x^2}}[/MATH]So I know that y is either positive or negative because of f(x). So the first [MATH]\pm[/MATH] gives me one option, not sure how I deal with the [MATH]\pm[/MATH] under the square roots. can both be valid depending on if my inverse is >0 or 0<?
[/QUOTE]
[MATH]u^2 + 2u - x^2 = 0 \implies u^2 + 2u = x^2 \implies u^2 + 2u + 1 = x^2 + 1 \implies \\ (u +1)^2 = x^2 + 1 \implies u + 1 = \pm \sqrt{x^2 + 1} \implies y^2 = \pm \sqrt{x^2 + 1} - 1.[/MATH]Let's start by getting the correct expression that is equal to y2. Calculus will teach you algebra.

[MATH]x^2 + 1 \ge 1 \implies \sqrt{x^2 + 1} \ge 1 \implies -\sqrt{x^2 + 1} \le -1 \implies \\ -\sqrt{x^2 + 1} - 1 \le -2 \implies y^2 < 0.[/MATH]So the negative option is impossible.

[MATH]\therefore y^2 = \sqrt{x^2 + 1} - 1 \implies y = \pm \sqrt{\sqrt{x^2 + 1} - 1}.[/MATH]
Now think about the original function. When x is negative, what is the sign of y? When x is zero, what is y? When x is positive, what is the sign of y?[/MATH]
 
If x is negative, so too is y and if x is positive, so too is y and whenever x is = 0 so is y.

So what is my domain and range for the function?
I guess my domain is: [MATH]x \in \mathbb{R} [/MATH] and my range: [MATH][0,\infty )[/MATH] since x=0 gives me y=0 which is the smallest value in my range.
[MATH]f^-1(x)=\pm \sqrt{-1+ \sqrt{(-1)^2+y^2}}[/MATH]
 
Ah, no, what I said is incorrect, the range and domain is all Reals because f(x) is never undefined and so we can plug in any x and get any y.

Thanks everyone for the help!
 
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