Find the inverse of the function

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  • #1
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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.
 

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  • #2
lanedance
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don't bother replacing x with y

just start with [tex] y = -2 \sqrt{ 3x-1 } -5[/tex]

and rearrange for x(y) this is your inverse function
 
  • #3
HallsofIvy
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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
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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
HallsofIvy
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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 [itex]x+ 5= -2\sqrt{3y- 1}[/itex]. Now the right hand side is multiplied by -2. You remove that by doing the opposite: dividing by -2:

[tex]\frac{x+ 5}{-2}= \sqrt{3y- 1}[/tex]

Now you need to get rid of the square root. What is the opposite of taking the square root?
 
  • #6
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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
HallsofIvy
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Well, you don't need that "-2 squared" in the denominator- it's just 4.
y= (x+5)^2/12+ 1.
 
  • #8
Mentallic
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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:

[tex]3y-1=\frac{(y+5)^2}{(-2)^2}[/tex]

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

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

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

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