# Functions with domains

I have the following question:

$$3$$. The functions $$f$$ and $$g$$ are defined with their respective domains by: $$f(x)=x^3,\quad x\in\mathbb{R}$$ $$g(x)=\frac{1}{x-3},\quad x\in\mathbb{R},\:\:x\ne3$$ a)$$\quad\:\:$$State the range of $$f$$.

b)$$\quad\:\:$$i)$$\quad\:\:$$ Find $$fg(x)$$.

$$\qquad\,\,$$ii)$$\quad\:$$ Solve the equation $$fg(x)=64$$.

c)$$\quad\:\:$$i)$$\quad\:\:$$ The inverse of $$g(x)$$ is $$g^{-1}(x)$$. Find $$g^{-1}(x)$$.

$$\qquad\,\,$$ii)$$\quad\:$$ State the range of $$g^{-1}(x)$$.

Here is my attempt. Are my answers correct?

Let $$f(x)=x^3,\:x\in\mathbb{R}$$ and $$g(x)=\dfrac{1}{x-3},\:x\in\mathbb{R},\:x\ne3$$.

a)$$\quad\:\:$$ The set $$\{f(x)\mid x\in\mathbb{R}\}$$ is called the range of $$f$$, and it is equal to $$\mathbb{R}$$.

b)$$\quad\:\:$$ $$f(x)=x^3$$ is a polynomial and $$g(x)=\dfrac{1}{x-3}$$.

$$\qquad\,\,$$ i)$$\quad\:\:$$ $$f(g(x))=f\left(\dfrac{1}{x-3}\right)=\left(\dfrac{1}{x-3}\right)^3=\dfrac{1}{(x-3)^3}$$

$$\qquad\,\,$$ii)$$\quad\:$$ $$f(g(x))=64$$

$$\qquad\qquad\dfrac{1}{(x-3)^3}=64$$

$$\qquad\qquad\quad\;\dfrac{1}{x-3}=4$$

$$\qquad\qquad\quad\:\,\,x-3=\dfrac{1}{4}$$

$$\qquad\qquad\:\:4(x-3)=1$$

$$\qquad\qquad\:\:\,4x-12=1$$

$$\hspace{31mm} 4x=13$$

$$\hspace{33mm} x=\dfrac{13}{4}$$

c)$$\quad\:\:\:$$i)$$\quad\:\:$$ Let $$g(x)=y$$. Then $$x=g^{-1}(y)$$.

$$\hspace{24mm} \dfrac{1}{x-3}=y$$

$$\hspace{26mm} x-3=\dfrac{1}{y}$$

$$\hspace{33.5mm} x=\dfrac{1}{y}+3$$

$$\hspace{33.5mm} x=\dfrac{1+3y}{y}$$

$$\hspace{24mm} g^{-1}(y)=\dfrac{1+3y}{y}$$

$$\hspace{21mm}$$ Now we can change variables:

$$\hspace{24mm} g^{-1}(x)=\dfrac{1+3x}{x}$$

$$\qquad\,\,$$ii)$$\quad\:$$ The set $$\{g^{-1}(x)\mid x\in\mathbb{R},\:x\ne0\}$$ is called the range of $$g^{-1}(x)$$, and it is equal to the $$\hspace{20mm}$$ domain of $$g(x)$$; i.e., the range of $$g^{-1}(x)$$ is $$\mathbb{R}\setminus \{3\}$$.

If part b) had said $$f(g(x))$$, I would be inclined to say your answers are all correct. In fact, it says $$fg(x)$$, which is usually taken to mean $$f(x)\times g(x)$$ and is sometimes written as $$(f\times g)(x)$$. In this case, $$fg(x)=\dfrac{x^3}{x-3}$$, and the resulting equation, $$fg(x)=64$$, doesn't look that difficult to solve. The rest of your answers are exactly right, though!