Piecewise Defined Function on Real Numbers
Question: If a function
\(f : \mathbb{R} \to \mathbb{R}\)
is defined by
$$
f(x)=
\begin{cases}
3x-2, & x<0 \\
1, & x=0 \\
4x+1, & x>0
\end{cases}
$$
Solution
The given function is a piecewise defined function. Different algebraic expressions are used for different values of \(x\).
The function is defined as:
$$ f(x)= \begin{cases} 3x-2, & x<0 \\ 1, & x=0 \\ 4x+1, & x>0 \end{cases} $$
This means:
- If \(x\) is negative, use \(f(x)=3x-2\).
- If \(x=0\), then \(f(0)=1\).
- If \(x\) is positive, use \(f(x)=4x+1\).
Examples
For \(x=-2\):
$$ f(-2)=3(-2)-2=-6-2=-8 $$
For \(x=0\):
$$ f(0)=1 $$
For \(x=3\):
$$ f(3)=4(3)+1=12+1=13 $$
Hence, the value of the function depends on the interval in which \(x\) lies.