Inverse Q (1 Viewer)

[azureus88]

how do you find the inverse of y=x^3 +3x. this is my working so far, but its wrong when you graph it...

x=y^3 +3y
x=y(y^2 +3)
xy=y^2 +3
y^2 - xy +3 = 0
y^2 - xy = -3
y^2 - xy + (x/2)^2 = -3 + (x/2)^2
(y - x/2)^2 = [(x^2 - 12)/4]
y - x/2 = +or- sqrt[(x^2 - 12)/4]
y = x/2 +or- sqrt[(x^2 - 12)/4]
y = 1/2[x +or- sqrt(x^2 - 12)]

the domain/range for original function is all real x but for my inverse,
(x^2 - 12) has to be greater than 0. can someone correct any mistakes ive done or show me an alternate way of doin it. thanks

 

[vds700]

how do you find the inverse of y=x^3 +3x. this is my working so far, but its wrong when you graph it...

x=y^3 +3y
x=y(y^2 +3)
xy=y^2 +3
y^2 - xy +3 = 0
y^2 - xy = -3
y^2 - xy + (x/2)^2 = -3 + (x/2)^2
(y - x/2)^2 = [(x^2 - 12)/4]
y - x/2 = +or- sqrt[(x^2 - 12)/4]
y = x/2 +or- sqrt[(x^2 - 12)/4]
y = 1/2[x +or- sqrt(x^2 - 12)]

the domain/range for original function is all real x but for my inverse,
(x^2 - 12) has to be greater than 0. can someone correct any mistakes ive done or show me an alternate way of doin it. thanks

u made a mistake in the bold line.

Also u have to consider that it is not a one to one function, so the u need to restrict the domian in order for it to have an inverse function.

atm i cant figure out how to express y in terms of x

 

[lyounamu]

how do you find the inverse of y=x^3 +3x. this is my working so far, but its wrong when you graph it...

x=y^3 +3y
x=y(y^2 +3)
xy=y^2 +3
y^2 - xy +3 = 0
y^2 - xy = -3
y^2 - xy + (x/2)^2 = -3 + (x/2)^2
(y - x/2)^2 = [(x^2 - 12)/4]
y - x/2 = +or- sqrt[(x^2 - 12)/4]
y = x/2 +or- sqrt[(x^2 - 12)/4]
y = 1/2[x +or- sqrt(x^2 - 12)]

the domain/range for original function is all real x but for my inverse,
(x^2 - 12) has to be greater than 0. can someone correct any mistakes ive done or show me an alternate way of doin it. thanks

see what you did there.

 

[azureus88]

ah...i see what i did wrong. its meant to be x/y=y^2 + 3

by the way, the original function is one-one

 

[vds700]

ah...i see what i did wrong. its meant to be x/y=y^2 + 3

by the way, the original function is one-one

did u sketch it correctly?

A one-one function means if u draw a horizontal line at any point, it will cut the curve at ONE point only.

Clearly this is not

EDIT: Actually, you're right,i sketched iut wrong, my bad.

 

[azureus88]

still cant seem to figure it out, cant get rid of the y^3

 

[lyounamu]

still cant seem to figure it out, cant get rid of the y^3

I cannot figure that part either. I actually went as far as using log and everything but no. Completing the square doesn't work either. May be we have to use something else...

 

[azureus88]

u reckon finding the derivative function, work out the inverse of that, then integrating, would work?

 

[lyounamu]

u reckon finding the derivative function, work out the inverse of that, then integrating, would work?

Yeah, it should work.

 

[azureus88]

tired it out but then wat are you supposed to sub in for the constant term?

 

[lyounamu]

tired it out but then wat are you supposed to sub in for the constant term?

You just don't. Work that out and change that back.

But to be honest, I don't really know how to do this particular question. It just wasted 20 minutes of my precious weekends.

 

[Iruka]

Well, the problem is not the constant term - you know that the original function passes through (0,0), so the inverse must pass through the origin also.

The problem is that you end up with gradient in terms of the wrong variable, and it is not clear how to substitute back into the correct one.

Unless there is some trick that I have missed, this is a pretty horrible question. I solved it by taking the OP's first line of working out

x=y^3+3y

and treating is as a cubic equation in y with x as the constant term. After about a page of working out, I found that the inverse is

y = cuberroot[(x+sqrt(x^2+4))/2] + cuberroot[(x-sqrt(x^2+4))/2]. (Looks horrible just writing it out!)

So basically, I think you have to know how to solve a cubic equation before you can get the inverse.

 

[azureus88]

Well, the problem is not the constant term - you know that the original function passes through (0,0), so the inverse must pass through the origin also.

The problem is that you end up with gradient in terms of the wrong variable, and it is not clear how to substitute back into the correct one.

Unless there is some trick that I have missed, this is a pretty horrible question. I solved it by taking the OP's first line of working out

x=y^3+3y

and treating is as a cubic equation in y with x as the constant term. After about a page of working out, I found that the inverse is

y = cuberroot[(x+sqrt(x^2+4))/2] + cuberroot[(x-sqrt(x^2+4))/2]. (Looks horrible just writing it out!)

So basically, I think you have to know how to solve a cubic equation before you can get the inverse.

can u briefly run over how you solved the cubic plz

 

[Iruka]

Well, this is my solution. It isn't exactly what I would call brief.

View attachment 17405

 

[azureus88]

Well, this is my solution. It isn't exactly what I would call brief.

View attachment 17405

woah shit, thats one epic proof! nice work dude. btw you in yr12?

 

[Iruka]

No. I'm a maths teacher.