# Composition of Functions

• MHB
• Ebone_Love

#### Ebone_Love

g(f(x))=$$g(f(x))=2/(1/x-4)+4$$

g(f(x))=$$g(f(x))=2/(1/x-4)+4$$
What is your question? If it's to find a possible combination of f(x) and g(x), then
$$\displaystyle f(x) = \dfrac{1}{x} - 4$$

and
$$\displaystyle g(x) = \dfrac{2}{x} + 4$$
will do the trick.

-Dan

Thank you, but I already know f(x) and g(x). I am trying to solve the equation for $$g(f(x))$$

I need help solving the written part of the question in the attached photo.

#### Attachments

• 20210627_212029.jpg
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You have g(f(x)). What are you trying to "solve?" Are you trying to graph it? Make a table?

$$\displaystyle \begin{array}{ l c r } x & f(x) & g(x) \\ -2 & -1/6 & -8 \\ -1 & -1/5 & -6 \\ 0 & -1/4 & -4 \\ 1 & -1/3 & -2 \\ 2 & -1/2 & 0 \\ \end{array}$$

Also: g(f(x)) = 2/( 1/(x - 4) ) + 4. You need an extra set of parenthesis.

-Dan

I am so sorry I forgot what I wanted to say , which was that I wanted to simplify g(f(x)).

Then you have to put $f(x)$ in $g(f(x))$ and solve. Considering the picture you have sent $f(x) =$ $1 \over (x-4)$ and $g(x)=$ $2 \over x$ $+4$
so we have $g(f(x))= 2(x-4) + 4 = 2x + 4$

Thank you topsquark.

g(f(x))=$$g(f(x))=2/(1/x-4)+4$$
$\frac{2}{\frac{1}{x}- 4}+ 4= \frac{2}{\frac{1}{x}- \frac{4x}{x}}+ 4= \frac{2}{\frac{1-4x}{x}}+ 4= \frac{2x}{1- 4x}+ 4$.

That would satisfy me but you could continue as
$\frac{2x}{1- 4x}+ \frac{4(1- 4x)}{1- 4x}= \frac{2x+ 4- 16x}{1- 4x}= \frac{4- 14x}{1- 4x}$.

Then you have to put $f(x)$ in $g(f(x))$ and solve. Considering the picture you have sent $f(x) =$ $1 \over (x-4)$ and $g(x)=$ $2 \over x$ $+4$
so we have $g(f(x))= 2(x-4) + 4 = 2x + 4$
No, 2(x- 4)+ 4= 2x- 8+ 4= 2x- 4.