Find the inverse of the function

1. Mar 6, 2009

HOFBrINCl

How would you find the inverse for this function: f(x)= -2 (squareroot 3x-1)-5, x greater than or equal to one third. I tried doing this through algebra (replacing x with y) but didn't get all the way.

2. Mar 6, 2009

lanedance

don't bother replacing x with y

just start with $$y = -2 \sqrt{ 3x-1 } -5$$

and rearrange for x(y) this is your inverse function

3. Mar 6, 2009

HallsofIvy

In other words, lanedance is telling you to do exactly what you say you did. How about showing your work so we can see where your problem is?

4. Mar 6, 2009

HOFBrINCl

So far I have x=-2(squaroot 3y -1) -5. Then you can make the 5 jump over, and that's as far as I've gotten. What's the next step here?

5. Mar 6, 2009

HallsofIvy

It's annoying enough when people say "move the 5 to the other side of the equation" but "make the 5 jump over" is just silly! You ADD 5 to both sides of the equation getting $x+ 5= -2\sqrt{3y- 1}$. Now the right hand side is multiplied by -2. You remove that by doing the opposite: dividing by -2:

$$\frac{x+ 5}{-2}= \sqrt{3y- 1}$$

Now you need to get rid of the square root. What is the opposite of taking the square root?

6. Mar 6, 2009

HOFBrINCl

haha, I've always said "make this or that jump over"! let's see, now I've come up with the complicated looking answer: y= (x+5/-2) squared, divided by 3 and +1. just looking at this it doesn't seem right, but I don't think I missed a step here.

7. Mar 6, 2009

HallsofIvy

Well, you don't need that "-2 squared" in the denominator- it's just 4.
y= (x+5)^2/12+ 1.

8. Mar 6, 2009

Mentallic

Be careful when manipulating equations in ways other than adding/subtracting. There is no jumping, leaping, or any other form of hopping in that sense :tongue2:

So you squared both sides:

$$3y-1=\frac{(y+5)^2}{(-2)^2}$$

Now, when dividing, be sure to divide EVERY term on both sides of the equation.

e.g. $$ax+b=c$$
solving for x: divide by a: $$x+\frac{b}{a}=\frac{c}{a}$$
and now isolate x by 'jumping' that 'thing' to the other side: $$x=\frac{c}{a}-\frac{b}{a}=\frac{c-b}{a}$$

It can also be done like this:
$$ax+b=c$$
$$ax=c-b$$
$$x=\frac{c-b}{a}$$

ok. See if you can apply this to the question now

Last edited: Mar 6, 2009