The Product of Functions

Operations on functions are performed in the same way as operations on numbers are performed.

The domain of the product of functions corresponds to the intersection of the domains of the functions in question. If there is a denominator, the restrictions on it must be excluded from the final domain.

Function |s| is defined by |s(x)=\dfrac{1}{(x^{2}-1)}| and function |t| is defined by |t(x)=x^{2}-x.| Multiplying the functions will result in: ||\begin{align}(s\times t)(x) &= s(x)\times t(x) \\ &=\dfrac{1}{(x^{2}-1)}\times (x^{2}-x) \\ &=\dfrac{1}{(x+1)(x-1)}\times (x)(x-1) \\ &= \dfrac{(x)\cancel{(x-1)}}{(x+1)\cancel{(x-1)}} \\ &= \dfrac{x}{x+1} \\ &=\frac{-1}{x+1}+1 \end{align}||

The domain of function |s| corresponds to |\mathbb{R}\backslash \lbrace -1,1 \rbrace| and the domain of function |t| corresponds to |\mathbb{R}.| The domain of the function given by |s\times t| will correspond to the intersection of the two initial domains. Next, we add the restriction to the denominator before the simplification |x \neq \lbrace -1, 1 \rbrace.| The result is |\mathbb{R} \backslash \lbrace -1,1 \rbrace.|

Function |u| is defined by |u(x)=\dfrac{2x^{2}-1}{x+3}| and function |v| is defined by |v(x)=-1.| ||\begin{align} (u\times v)(x) &= u(x)\times v(x) \\ &=\dfrac{2x^{2}-1}{x+3}\times -1 \\ &=\frac{-2x^{2}+1}{x+3} \end{align}||

The domain of function |u| corresponds to |\mathbb{R} \backslash \lbrace -3 \rbrace| and the domain of function |v| corresponds to |\mathbb{R}|. The domain of the function given by |u\times v| will correspond to the intersection of the two initial domains, to which we must add the restriction to the denominator |x \neq -3.| Therefore, this function’s domain will be |\mathbb{R} \backslash \lbrace -3 \rbrace.|

To find the product of functions in a graph, multiply the range of the first function by the range of the second function.

To produce the graph, create a table of values or use the peculiarities of the resulting function.

In the first example, a table of values of the functions |s(x)=\dfrac{1}{(x^{2}-1)},| |t(x)=x^{2}-x| and |(s\times t)(x)=\dfrac{x}{x+1},| will result in:

Since the resultant is a rational function, its peculiarities can be used to plot the equation.

||(s\times t)(x)=\dfrac{-1}{(x+1)}+1||

• There are two asymptotes: |x=-1| et |y =1.|

• Since |ab<0,| the function is located in the 2^nd and 4^th quadrant. The result is the following graph:

Do not forget the restriction of |x=1.| The other restriction corresponds to the asymptote of equation |x=-1.|