Thread: Inverse & Composition Function Problem

1. Inverse & Composition Function Problem

#1.)

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

I'm having trouble determining what $\displaystyle g(x)$ 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 $\displaystyle h(x$) I would have just decomposed $\displaystyle h(x)$ as $\displaystyle a(x) = 4x$ and $\displaystyle b(x) = x-1 <=> a(b(x)) = 4(x-1)$ - but I'm obviously forced to use the inside function $\displaystyle f(x) = x+3$.

#2.)

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

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.

2. Originally Posted by TaylorM0192
#1.)

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

I'm having trouble determining what $\displaystyle g(x)$ 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 $\displaystyle h(x$) I would have just decomposed $\displaystyle h(x)$ as $\displaystyle a(x) = 4x$ and $\displaystyle b(x) = x-1 <=> a(b(x)) = 4(x-1)$ - but I'm obviously forced to use the inside function $\displaystyle f(x) = x+3$.
try $\displaystyle g(x)=4x-8$

3. 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...

4. $\displaystyle g(x+3)= 4x-4$

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

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

Expand

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

Equate coefficents

$\displaystyle ax= 4x \implies a=4$

and

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

$\displaystyle g(x)= 4x-16$

5. 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 -

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

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

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

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

Thanks ...

7. Originally Posted by TaylorM0192

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

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

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

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

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