Find the inverse of the function

[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.

 

[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

 

[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?

 

[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?

 

[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?

 

[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.

 

[HallsofIvy]

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

 

[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

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