Expert: Paul Klarreich - 11/9/2006

Question
i cannot for the life of me figure out how to do this problem! i know the concept and i've done others like it but this one just isn't working for me...

Find the inverse of f(x)=x2+4x=4. restrict the domain,if necessary. (that two following the x is to square it)
thanks!!

Answer
Questioner:  missy
Category:  Advanced Math
 
Subject:  Inverses of functions.<< I CHANGED IT.
Question:  i cannot for the life of me figure out how to do this problem! i know the concept and i've done others like it but this one just isn't working for me...

Find the inverse of f(x)=x2+4x=4.

>> You meant  f(x) = x^2 + 4x - 4, I presume.  If it was supposed to be + 4 at the end, you can make the appropriate changes yourself -- it won't be hard.

restrict the domain,if necessary.

>> It will be.

thanks!
..........................................
Hi, Missy,

Here's the scoop on getting the inverse of a function.  If the graph has a turning point, (use your graphing calculator to see it) then you ARE going to have more than one value of x giving the same value of y.

If you invert, then you will have more than one value of y for a given x, and THAT IS A NO NO.

Now the graph of  f(x) = x^2 + 4x - 4  is a parabola, which I am sure you knew, and so it has a turning point, which is, in fact, at  x = -2.

So if you take a portion of the graph (of the domain?) which includes stuff on both sides of that turning point, you will not get an inverse for your function.

Solution:  Take only the part of the domain that is on one side of the T.P.  Since the T.P. is at x = -2, take only values of x >= -2  (or only values of x <= -2)

So our domain restriction makes:

f(x) =  x^2 + 4x - 4,  and  x >= -2

Now we would like to compute the inverse.  To make an inverse, the steps are:

A. Write  y = f(x).
B. Exchange the x and y symbols everywhere.
C. Solve for y in terms of x.
D. Write  f^-1(x) = y

y =  x^2 + 4x - 4,  and  x >= -2    << writing y = f(x).  Note the x>=-2 is part of it.

x =  y^2 + 4y - 4,  and  y >= -2     << exchange. See how it becomes  y>=-2.  When I said everywhere, I MEANT EVERYWHERE.

Solve.  This may take some effort. My favorite method for quadratics is completing the square.  With some practice, you may come to love it as I do.

y^2 + 4y - 4 = x         and...

y^2 + 4y     = x + 4          and...

y^2 + 4y + 4 = x + 4 + 4     and...

(y + 2)^2 = x + 8     and...

y + 2 =  +- sqrt(x + 8)     and...

y = -2 +- sqrt(x + 8)  and  y >= -2

Since y must be >= -2, we want the positive sqrt part of that solution:

y = -2 + sqrt(x + 8)

Finally, we can write:

f^-1(x) = -2 + sqrt(x + 8)

.............................
Another note about that:  You see that when we solve a quadratic equation we get that pesky '+-' in there, which we don't want.  Therefore, we need to make some restriction that forces either '+' or '-'.  We could have worked it backwards like this:

We got  y = -2 +- sqrt(x + 8)   after solving for y.

We must choose either  + or -.  Let's say we choose '+'.  Then we will say:

y = -2 + sqrt(x + 8)  is the function, and since  sqrt(..) >= 0, y >= -2.

If y >= -2 in our inverse, we must have had  x >= -2  in our original function, so that is our domain restriction.

Not happy with the choice of '-'?  You say you're not satisfied?  You say you want more for your money?  Tell you what I'm gonna do.  I'll pick the '-' in the formula.

y = -2 - sqrt(x + 8)  is the function, and since  sqrt(..) >= 0, y <= -2.

[Just work it out for yourself -- if you subtract a positive number, your answer gets smaller.]

If y <= -2 in our inverse, we must have had  x <= -2  in our original function, so that is our domain restriction.

Remember, we said we could pick either side of that turning point.