Practice questions

$\dfrac{f}{g}=\dfrac{f(x)}{g(x)}$

 $=\dfrac{x^2-16}{x^2-10x+25}\div \dfrac{3x-12}{x^2-3x-10}$

Change the division into multiplication by taking the reciprocal of the rational expression that follows the $\div$ symbol:

 $=\dfrac{x^2-16}{x^2-10x+25}\cdot \dfrac{x^2-3x-10}{3x-12}$

Factor the expressions in the numerator and the denominator:

 $=\dfrac{(x+4)(x-4)}{(x-5)(x-5)}\cdot \dfrac{(x-5)(x+2)}{3(x-4)}$

Cancelling that can be cancelled:

 $=\dfrac{(x+4)(\cancel{x-4})}{(x-5)(\cancel{x-5})}\cdot \dfrac{(\cancel{x-5})(x+2)}{3(\cancel{x-4})}$

 $=\dfrac{(x+4)(x+2)}{3(x-5)}$

$(f-g)(x)=f(x)-g(x)$

Substituting $f(x)$ and $g(x)$:

 $=\dfrac{5ab}{a^2-b^2}-\dfrac{a-b}{a+b}$

Factoring $a^2-b^2=(a+b)(a-b)$:

 $=\dfrac{5ab}{(a+b)(a-b)}-\dfrac{a-b}{a+b}$

LCD$=(a+b)(a-b)$

Divide the LCD by the denominator of each term and multiply that. And put the LCD as the common denominator:

 $=\dfrac{5ab-(a-b)(a-b)}{(a+b)(a-b)}$

 $=\dfrac{5ab-(a-b)^2}{(a+b)(a-b)}$

Expanding, $(a-b)^2=a^2-2ab+b^2$:

 $=\dfrac{5ab-(a^2-2ab+b^2)}{(a+b)(a-b)}$

 $=\dfrac{5ab-a^2+2ab-b^2}{(a+b)(a-b)}$

 $=\dfrac{7ab-a^2-b^2}{(a+b)(a-b)}$

$(f\cdot g)(-3)=f(-3)\cdot g(-3)$

You need to find $f(-3)$ and $g(-3)$ and multiply them.

$f(-3)=\dfrac{3(-3)}{6(-3)^2-13(-3)-5}$

  $=\dfrac{-9}{54+39-5}$

  $=\dfrac{-9}{88}$

$g(-3)=4(-3)-10$

 $=-12-10$

 $=-22$

$(f\cdot g)(-3)=\dfrac{-9}{88}\cdot (-22)$

    $=\dfrac{9}{4}$

The given function is a one-to-one function.

To find the inverse, $g^{-1}$

1. Replace, $g(x)$ by $y$:

$y=\sqrt{x-3}$

2. Switch $x$ and $y$:

$x=\sqrt{y-3}$

3. Solve for $y$:

The variable $y$ is inside the square root. To solve for $y$, you need to remove the square root by squaring on both sides:

$x^2=(\sqrt{y-3})^2$

$x^2=y-3$

Solving for $y$:

$y=x^2+3$

This $y$ is the inverse.

Thus, $g^{-1}(x)=x^2+3$

$(f\circ g)(x)=f(g(x))$

Take the function $f(x)$ and replace the $x$ with the function $g(x)$. Since $g(x)=\dfrac{2}{x}$. So, you need to replace $x$ in $f(x)$ by $\dfrac{2}{x}$

$(f\circ g)(x)=4\left(\dfrac{2}{x}\right)^2-1$

 $=4\left(\dfrac{2^2}{x^2}\right)-1$

 $=4\left(\dfrac{4}{x^2}\right)-1$

 $=\dfrac{16}{x^2}-1$

$(g\circ f)(x)=g(f(x))$

Take the function $g(x)$ and replace the $x$ with the function $f(x)$. Since $f(x)=4x^2-1$. So, you need to replace $x$ by $4x^2-1$.

$(g\circ f)(x)=\dfrac{2}{4x^2-1}$