Inverse Functions

**Macleef** · Jan 10th 2008, 03:03 PM

Find the inverses of the following functions

o) $y = 3(x - 1)^2, x \geq 1$

$x = 3(y - 1)^2$

$x = 3(y - 1)(y - 1)$

$x = 3(y^2 - 2y + 1)$

$x = 3y^2 - 6y + 3$

$x - 3 = 3y^2 - 6y$

$x - 3 = 3y(y - 2)$

$\frac {x - 3}{y - 2} = 3y$

$\frac{\frac {x - 3}{y - 2}}{3} = y$

$(3)\frac {x - 3}{y - 2} = y$

$\frac{3x - 9}{y - 2} = y$

$3x - 9 = \frac {y}{y - 2}$

$3x - 9 = 1 - \frac {y}{2}$

$3x - 9 - 1= \frac {y}{2}$

$3x - 10 = \frac {y}{2}$

$\frac {3x - 10}{2} = y$

I'm way off. . .

And textbook answer is $y = 1 + \sqrt \frac {x}{3}, x \geq 0$

**Jhevon** · Jan 10th 2008, 03:07 PM

Originally Posted by Macleef

Find the inverses of the following functions

o) $y = 3(x - 1)^2, x \geq 1$

$x = 3(y - 1)^2$

$x = 3(y - 1)(y - 1)$

$x = 3(y^2 - 2y + 1)$

$x = 3y^2 - 6y + 3$

$x - 3 = 3y^2 - 6y$

$x - 3 = 3y(y - 2)$

$\frac {x - 3}{y - 2} = 3y$

$\frac{\frac {x - 3}{y - 2}}{3} = y$

$(3)\frac {x - 3}{y - 2} = y$

$\frac{3x - 9}{y - 2} = y$

$3x - 9 = \frac {y}{y - 2}$

$3x - 9 = 1 - \frac {y}{2}$

$3x - 9 - 1= \frac {y}{2}$

$3x - 10 = \frac {y}{2}$

$\frac {3x - 10}{2} = y$

I'm way off. . .

And textbook answer is $y = 1 + \sqrt \frac {x}{3}, x \geq 0$

here's the trick. think of x as if it was just a regular number. then when you expand everything, and move everything to one side, you'd realize that the expression you have is quadratic in y. just use the quadratic formula and you will get the answer

**Macleef** · Jan 10th 2008, 03:12 PM

Originally Posted by Jhevon

here's the trick. think of x as if it was just a regular number. then when you expand everything, and move everything to one side, you'd realize that the expression you have is quadratic in y. just use the quadratic formula and you will get the answer

I solve the expression and got. . .

x = (3y - 3)(y - 1)

and I don't know what to do now

**Jhevon** · Jan 10th 2008, 03:16 PM

Originally Posted by Macleef

I solve the expression and got. . .

x = (3y - 3)(y - 1)

and I don't know what to do now

as i said, expand everything. you want to get it in the form $ay^2 + by + c = 0$
and the expression for c will have x in it

so you just use the quadratic formula: $y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

**Macleef** · Jan 10th 2008, 03:24 PM

Originally Posted by Jhevon

as i said, expand everything. you want to get it in the form $ay^2 + by + c = 0$
and the expression for c will have x in it

so you just use the quadratic formula: $y = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$

"the expression for c will have x in it" . . .I don't get this part. . .

and when I tried solving this equation with the quadratic formula, I ended up with. . .

$x = \frac {6 \pm \sqrt{0}}{6}$

**Jhevon** · Jan 10th 2008, 03:33 PM

Originally Posted by Macleef

"the expression for c will have x in it" . . .I don't get this part. . .

and when I tried solving this equation with the quadratic formula, I ended up with. . .

$x = \frac {6 \pm \sqrt{0}}{6}$

where did the y go?

ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get

**Macleef** · Jan 10th 2008, 03:37 PM

Originally Posted by Jhevon

where did the y go?

ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get

x = 3(y - 1)(y - 1)
0 = 3(y - 1)(y - 1) - x

**Jhevon** · Jan 10th 2008, 03:46 PM

Originally Posted by Macleef

x = 3(y - 1)(y - 1)
0 = 3(y - 1)(y - 1) - x

expand means there should be no brackets left

**Macleef** · Jan 10th 2008, 03:51 PM

Originally Posted by Jhevon

expand means there should be no brackets left

Oh. . .

Is the following what you mean. . .?

$0 = 3y^2 - 6y + 3 - x$

**badgerigar** · Jan 10th 2008, 03:53 PM

Quote:
Originally Posted by Jhevon View Post
where did the y go?

ok, let's take this one step at a time. first step, expand everything as much as possible and bring everything to one side. type what you get
x = 3(y - 1)(y - 1)
0 = 3(y - 1)(y - 1) - x

This will work and is a correct first step on a way of solving this. The next step of this method is to expand the brackets. (EDIT: sorry didn't see new posts. You already got this part)

However, I think the problem is more sensibly done this way:

$x = 3(y-1)^2$
$<br /> \frac x3 = (y-1)^2$
$<br /> \pm \sqrt {\frac x3} = y-1$
$y = 1 \pm \sqrt {\frac x3}$

**Jhevon** · Jan 10th 2008, 03:56 PM

Originally Posted by badgerigar

This will work and is a correct first step on a way of solving this. The next step of this method is to expand the brackets. (EDIT: sorry didn't see new posts. You already got this part)

However, I think the problem is more sensibly done this way:

$x = 3(y-1)^2$
$<br /> \frac x3 = (y-1)^2$
$<br /> \pm \sqrt {\frac x3} = y-1$
$y = 1 \pm \sqrt {\frac x3}$

well, that was easy

we need a staples easy button around here

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