0
$\begingroup$

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 $\mathbb{R}\setminus \{3\}$.

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.