Functions defined by parts

A function defined by parts is a function whose analytic expression is not unique but rather it depends on the value of the independent variable.

The function $$f(x)=\left\{\begin{array}{rcl} -x-1 & \mbox{ if } & x \leq -3 \\ 3 & \mbox{ if } & -1 < x < 1 \\ x-2 & \mbox{ if } & x\geq 1 \end{array}\right.=\left\{\begin{array}{rcl} -x-1 & \mbox{ if } & x\in (-\infty,3] \\ 3 & \mbox{ if } & x\in (-1,1)\\ x-2 & \mbox{ if } & x\in [1, +\infty]\end{array}\right.$$ It is a function defined by parts.

To find the image of an element $x$ we need to take into account at what interval it belongs to and replace it in the analytic expression corresponding to this interval.

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. $$Dom(f)=(-\infty,-3] \cup (-1,1)\cup [1,+\infty)=(-\infty,-3]\cup (-1,+ \infty)$$ If we look now at the graph of the previous function, we can observe that $Im (f)=[-1,+\infty)$:

Let's see some examples of functions defined by parts:

Consider the function $\displaystyle f(x)=\left\{\begin{array}{rcl} 1 & \mbox{ if } & x\leq 2 \\ 2 & \mbox{ if } & x > 2\end{array}\right.$.

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

The graphic representation would be:

Consider the function $\displaystyle f(x)=\left\{\begin{array}{rcl} -x & \mbox{ if } & x\leq -1 \\ x^2 & \mbox{ if } & -1 < x < 1 \\ x & \mbox{ if } & x \geq 1 \end{array}\right.$

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.

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

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

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

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

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

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.