SOLUTION: If {{{g(x)=x^2+4x}}} find {{{g^-1(5)}}} I know you need to solve for x but I can't remember how. {{{y=x^2+4x}}} {{{y=x(x+4)}}} {{{y/(x+4) = x}}} but I still have

Algebra ->  Rational-functions -> SOLUTION: If {{{g(x)=x^2+4x}}} find {{{g^-1(5)}}} I know you need to solve for x but I can't remember how. {{{y=x^2+4x}}} {{{y=x(x+4)}}} {{{y/(x+4) = x}}} but I still have       Log On


   

Question 125918This question is from textbook Elementary Functions
: If g%28x%29=x%5E2%2B4x find g%5E-1%285%29
I know you need to solve for x but I can't remember how.
y=x%5E2%2B4x
y=x%28x%2B4%29
y%2F%28x%2B4%29+=+x
but I still have an x with the y??
Please help. I know the solution should be x = some equation where the equation cannot also have an x in it.
thank you This question is from textbook Elementary Functions

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
y=x%5E2%2B4x
y=x%28x%2B4%29
y%2F%28x%2B4%29+=+x
What you have is a relation between
2 variables. You can't "solve" any more
than you did when you got y+=+x%28x+%2B+4%29
If I say y+=+10, then there is a
corresponding value for x, but normally
y depends on whatever x is.
You can find the roots of y=x%5E2%2B4x
by setting y+=+0. You can see in
y=x%28x%2B4%29 than x+=+0 and x+=+-4
will make y+=+0