Functions defined by parts

Example

The function $$f(x)=\{\begin{array}{rcl}-x-1& \text{ if }& x\le -3\\ 3& \text{ if }& -1<x<1\\ x-2& \text{ if }& x\ge 1\end{array}=\{\begin{array}{rcl}-x-1& \text{ if }& x\in (-\mathrm{\infty },3]\\ 3& \text{ if }& x\in (-1,1)\\ x-2& \text{ if }& x\in [1,+\mathrm{\infty }]\end{array}$$ It is a function defined by parts.

Example

In the previous case for example,

• if $x=-4$, we substitute in $f(x)=-x-1$ and obtain $f(-4)=3$
• if $x=-2$, the image is not defined since $-2$ does not belong to any interval of the function.
• if $x=0.5$, we substitute in $f(x)=3$ obtaining $f(0.5)=3$
• if we $x=1$ substitute in $f(X)=x-2$ obtaining $f(1)=-1$

Since each of the expressions of the parts are defined in some domain, the domain of the function $f(x)$ is the union of the intervals of the parts of the function. $$Dom(f)=(-\mathrm{\infty },-3]\cup (-1,1)\cup [1,+\mathrm{\infty })=(-\mathrm{\infty },-3]\cup (-1,+\mathrm{\infty })$$ If we look now at the graph of the previous function, we can observe that $Im(f)=[-1,+\mathrm{\infty })$:

Example

Consider the function ${\displaystyle f(x)=\{\begin{array}{rcl}1& \text{ if }& x\le 2\\ 2& \text{ if }& x>2\end{array}}$.

Its graph is the union of the graphs of the functions $f(x)=1$ for $x\le 2$ and $f(x)=2$ for $x>2$.

The graphic representation would be:

Example

Consider the function ${\displaystyle f(x)=\{\begin{array}{rcl}-x& \text{ if }& x\le -1\\ x^2& \text{ if }& -1<x<1\\ x& \text{ if }& x\ge 1\end{array}}$

This time its graph will be the union of a straight line, a parable and another straight line, with each one defined where it is indicated in the definition of the function.

Example

${\displaystyle f(x)=\{\begin{array}{rcl}2x-1& \text{ if }& x<1\\ x+3& \text{ if }& x>1\end{array}}$

and if we want to evaluate at $x=-1$ in this example we will obtain:$f(-1)=f_1(-1)=2(-1)=2(-1)-1=-2-1=-3$

Example

${\displaystyle f(x)=\{\begin{array}{rcl}x-1& \text{ if }& x<-3\\ x^2+1& \text{ if }& -3\le <0\\ 3& \text{ if }& 0\le x\le 100\\ \ln (x+e^x)& \text{ if }x>100\end{array}}$

and if we want to evaluate at $x=-1$ in this example we will obtain: $f(-1)=f_2(-1)=(-1{)}^2+1=1+1=2$

Example

The following example is not a function by parts since the sets of the definition are not disjoint: $${\displaystyle f(x)=\{\begin{array}{rcl}x& \text{ if }& x<0\\ x+1& \text{ if }& -1<x<2\\ -3& \text{ if }x\ge 2\end{array}}$$

since for points in $(-1,0)$ it would be necessary to evaluate the function at $f_1(x)=x$ and at $f_2(x)=x+1$, and thus we would obtain two different values for only one point and this would not be a function.