Question

Combining Functions Find $f+g,f-g,fg$and $\frac{f}{g}$and their domains.

$f\left(x\right)=\frac{2}{x+1},g\left(x\right)=\frac{x}{x+1}$

Short answer

The required functions and their domains are:

The domain of$\left(f+g\right)\left(x\right)=\frac{2+x}{x+1}$ is$\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(f-g\right)\left(x\right)=\frac{2-x}{x+1}$ is$\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(fg\right)\left(x\right)=\frac{2x}{{\left(x+1\right)}^2}$ is$\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(\frac{f}{g}\right)\left(x\right)=\frac{2}{x}$ is $\left(-\infty ,-1\right)\cup \left(-1,0\right)\cup \left(0,\infty \right)$.

Step-by-step solution

– Domain of combined functions.

Domain of a function is the set of all possible input values.

If domains of f and g are A and B then,

• Domain of$f+g$ is $A\cap B$.
• Domain of$f-g$ is $A\cap B$.
• Domain of fg is $A\cap B$.
• Domain of$\frac{f}{g}$ is $\left\{x\in A\cap B|g\left(x\right)\ne 0\right\}$.

– Given information.

The given functions are:

$\begin{array}{c}f\left(x\right)=\frac{2}{x+1}\\ g\left(x\right)=\frac{x}{x+1}\end{array}$

The function$f\left(x\right)$ defined if$x+1\ne 0$ that $x\ne -1$. So, domain of f is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The function$g\left(x\right)$ defined if$x+1\ne 0$ that $x\ne -1$. So, domain of g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

Intersection of domains of f and g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

– Sum of given functions and its domain.

$\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\frac{2}{x+1}+\frac{x}{x+1}\\ =\frac{2+x}{x+1}\end{array}$

The intersection of domains of both functions f and g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$. So, the domain of ( f + g ) is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

– Difference of given functions and its domain.

$\begin{array}{c}\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\\ =\frac{2}{x+1}-\frac{x}{x+1}\\ =\frac{2-x}{x+1}\end{array}$

The intersection of domains of both functions f and g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$. So, the domain of ( f - g ) is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

– Product of given functions and its domain.

$\begin{array}{c}\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\\ =\frac{2}{x+1}\cdot \frac{x}{x+1}\\ =\frac{2x}{{\left(x+1\right)}^2}\end{array}$

The intersection of domains of both functionsf and g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$. So, the domain of ( fg ) is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

– Quotient of given functions and its domain.

$\begin{array}{c}\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ =\frac{\frac{2}{x+1}}{\frac{x}{x+1}}\\ =\frac{2}{x+1}\cdot \frac{x+1}{x}\\ =\frac{2}{x}\end{array}$

The intersection of domains of both functions f and g is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The function$g\left(x\right)=0$ if $\frac{x}{x+1}=0\Rightarrow x=0$.

So, the domain of$\left(\frac{f}{g}\right)$ is$\left(-\infty ,-1\right)\cup \left(-1,0\right)\cup \left(0,\infty \right)$.

Thus,the required functions are their domains are:

The domain of$\left(f+g\right)\left(x\right)=\frac{2+x}{x+1}$ is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(f-g\right)\left(x\right)=\frac{2-x}{x+1}$ is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(fg\right)\left(x\right)=\frac{2x}{{\left(x+1\right)}^2}$ is $\left(-\infty ,-1\right)\cup \left(-1,\infty \right)$.

The domain of$\left(\frac{f}{g}\right)\left(x\right)=\frac{2}{x}$ is $\left(-\infty ,-1\right)\cup \left(-1,0\right)\cup \left(0,\infty \right)$.