Question

A. Use composition to prove whether or not the functions are inverses of each other.

Answer 1

A. In a composition of two functions the first function is evaluated, and then the second function is evaluated on the result of the first function. In other word, you are going to evaluate the second function in the first function.

Remember that you can evaluate function at any number just replacing the variable in the function with the number. For example, let's evaluate our function $f(x)$ at $x=1$:

$f(x)=\frac{1}{x-3}$

$f(1)=\frac{1}{1-3}$

$f(1)=\frac{1}{-2}$

Similarly, to find the composition of $f(x)$ and$g(x)$, we are going to evaluate $f(x)$ at $g(x)$. In other words, we are going to replace $x$ in $f(x)$ with $\frac{3x+1}{x}$:

$f(x)=\frac{1}{x-3}$

$f(g(x) = f(\frac{3x+1}{x} ) = \frac{1}{\frac{3x+1}{x} -3}$

Remember that two functions are inverse if after simplifying their composition, we end up with just $x$. Let's simplify and see what happens.

$f(g(x)=\frac{1}{\frac{3x+1}{x} -3}$

$f(g(x)=\frac{1}{\frac{3x+1-3x}{x} }$

$f(g(x)=\frac{1}{\frac{1}{x} }$

$f(g(x)=x$

Now let's do the same for $g(f(x))$:

$g(\frac{1}{x-3} )=\frac{3(\frac{1}{x-3})+1}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{3}{x-3}+1}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{3+x-3}{x-3}}{x}$

$g(\frac{1}{x-3} )=\frac{\frac{x}{x-3}}{x}$

$g(\frac{1}{x-3} )=\frac{x}{x(x-3)}$

$g(f(x))=\frac{x}{x(x-3)}$

We can conclude that $g(x)$ is the inverse function of $f(x)$, but $f(x)$ is not the inverse function of $g(x)$.

B. The domain of a function is the set of all the possible values the independent variable can have. In other words, the domain are all the possible x-values of function.

Now, interval notation is a way to represent and interval using an ordered pair of numbers called the end points; we use brackets [ ] to indicate that the end points are included in the interval and parenthesis ( ) to indicate that they are excluded.

Notice that when $x=0$, $g(x)=\frac{3(0)+1}{0} =\frac{0}{0}$, so when $x=0$, $g(x)$ is not defined; therefore we have to exclude zero from the domain of $f(g(x))$.

We can conclude that the domain of the composite function $f(g(x))$ in interval notation is (-∞,0)U(0,∞)

Now let's do the same for $g(f(x))$.

Notice that the composition is not defined when its denominator equals zero, so we are going to set its denominator equal to zero to find the values we should exclude from its domain:

$x(x-3)=0$

$x=0$ and $x-3=0$

$x=0$ and $x=3$

Know we know that we need to exclude $x=0$ and $x=3$ from the domain of $g(f(x))$.

We can conclude that the domain of the composition function $g(f(x))$ is (-∞,0)U(0,3)U(3,∞)

Answer 2

Answer: Hello mate!

A composition of two functions is when a function is valuated in the other:

this is for f(x) and g(x); f(g(x))

now a the functions are inverse if: f(g(x)) = x  

in this case $f(x) = \frac{1}{x-3}$ and $g(x) = \frac{3x + 1}{x}$

then $f(g(x)) = \frac{1}{(\frac{3x + 1}{x} - 3) }$

now let's simplify it:

$f(g(x)) = \frac{1}{(\frac{3x + 1}{x} - 3) } = x\frac{1}{(3x + 1 - 3x) } = x\frac{1}{( 1 ) } = x$

Now let's see g(f(x)), we can suppose that this is also equal to x, but let's prove it:

$g(f(x)) = \frac{3(\frac{1}{x-3}) + 1 }{\frac{1}{x-3} } =(x-3)(3(\frac{1}{x-3}) + 1 ) = 3+x-3 =x$

as we expected, then f(x) and g(x) are inverses of each other.

b) the composition f(g(x)) = x

is true that the domain of f(x) is the set of all real numbers different to 3, and the domain of g(x) is the set of all real numbers except the 0 ( you can see the exceptions by loking when the denominator is equal to 0)

but in the composition there are not a problem with the denominator (the denominator is equal to 1 in this case)

so the domain of the composition is the set of all real numbers