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)$\quad\:$ $f(g(x))=64$






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

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

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!


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