Inverse & Composition Function Problem

**TaylorM0192** · February 2nd 2010, 06:29 PM

#1.)

Given $f(x) = x+3$ and $h(x) = 4x-4 <=> 4(x-1)$ ; find a function $g(x)$ such that $g(f(x)) = h(x) <=> g(x+3) = 4(x-1)$ .

I'm having trouble determining what $g(x)$ should be - I don't know the method to determine this. If I was just asked to find any two functions whos composition resulted in $h(x$ ) I would have just decomposed $h(x)$ as $a(x) = 4x$ and $b(x) = x-1 <=> a(b(x)) = 4(x-1)$ - but I'm obviously forced to use the inside function $f(x) = x+3$ .

#2.)

Given $f(x) = 3x + ln(x)$ find $f^-1(x)$ . In other words, solve for $y$ in $x = 3y + ln(y)$ .

I posted #2 in a previous thread, but I didn't really get the response I was looking for - I would appreciate the explanation for the answer if possible.

**pickslides** · February 2nd 2010, 07:22 PM

Originally Posted by TaylorM0192

#1.)

Given $f(x) = x+3$ and $h(x) = 4x-4 <=> 4(x-1)$ ; find a function $g(x)$ such that $g(f(x)) = h(x) <=> g(x+3) = 4(x-1)$ .

I'm having trouble determining what $g(x)$ should be - I don't know the method to determine this. If I was just asked to find any two functions whos composition resulted in $h(x$ ) I would have just decomposed $h(x)$ as $a(x) = 4x$ and $b(x) = x-1 <=> a(b(x)) = 4(x-1)$ - but I'm obviously forced to use the inside function $f(x) = x+3$ .

try $g(x)=4x-8$

**TaylorM0192** · February 2nd 2010, 07:28 PM

Your g(x) does not result in 4x-4 when composed with f(x) = x+3 as the inner function. However, thank you for pointing me in the right direction, since it appears g(x) = 4x-16 works.

But I still would like to know the steps to approach problems like these, since the method of trial and error in finding a correct function seems impractical later on.

And the 2nd question is still giving me problems...I simply don't know what to do with the logarithm when the variable I am solving for is in two separate terms...

**pickslides** · February 2nd 2010, 07:58 PM

$g(x+3)= 4x-4$

Lets say $g(x)=ax+b$

$a(x+3)+b= 4x-4$

Expand

$ax+3a+b= 4x-4$

Equate coefficents

$ax= 4x \implies a=4$

and

$3a+b= -4$ with $a=4$ gives $3\times 4+b= -4\implies b = -16$

$g(x)= 4x-16$

**TaylorM0192** · February 2nd 2010, 08:11 PM

Thanks bud, appreciate the explanation.

Any suggestions on the 2nd problem? I started by isolating the natural logarithm, them setting each term as an exponent with base 'e' (thus isolating 1 'y' by itself). But then the other 'y' becomes an exponent which can only be isolated by going back to the original equation and taking the natural logarithm of both sides...Whatever I try, I end up with ugly terms -

**TaylorM0192** · February 3rd 2010, 08:25 AM

I've gotten to this point in trying to find the inverse for $f(x)=ln(x)+3x$ :

$x=ln(y)+3y$
$ln(y)=x-3y$
$y=e^{x-3y}$

Is it even possible to isolate 'y' to a single term to create an inverse function?

The ultimate question is, what is $f^{-1}(3)$ of $f(x)=ln(x)+3x$ .

Thanks ...

**pickslides** · February 3rd 2010, 11:55 AM

Originally Posted by TaylorM0192

The ultimate question is, what is $f^{-1}(3)$ of $f(x)=ln(x)+3x$ .

Thanks ...

Thats better, giving a bit more information like that makes life easier.

consider $f(x)=\ln(x)+3x$

by inspection $f(1)=\ln(1)+3\times 1 = 0+3=3$

so we can say $f^{-1}(3)= 1$

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