Question

write your own continuous piecewise function with at least three different sub - functions. how did you ensure that the function was continuous?

Explanation:

Step1: Define the piece - wise function

Let \(f(x)=

$$\begin{cases}x + 1, &x<0\\x^{2}+1, &0\leq x<1\\2x, &x\geq1\end{cases}$$

\)

Step2: Check continuity at \(x = 0\)

Left - hand limit: \(\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{-}}(x + 1)=0 + 1=1\)
Right - hand limit: \(\lim_{x
ightarrow0^{+}}f(x)=\lim_{x
ightarrow0^{+}}(x^{2}+1)=0^{2}+1 = 1\)
\(f(0)=0^{2}+1 = 1\). Since \(\lim_{x
ightarrow0^{-}}f(x)=\lim_{x
ightarrow0^{+}}f(x)=f(0)\), the function is continuous at \(x = 0\).

Step3: Check continuity at \(x = 1\)

Left - hand limit: \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(x^{2}+1)=1^{2}+1=2\)
Right - hand limit: \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(2x)=2\times1 = 2\)
\(f(1)=2\times1=2\). Since \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)=f(1)\), the function is continuous at \(x = 1\).

Answer:

The piece - wise function \(f(x)=

$$\begin{cases}x + 1, &x<0\\x^{2}+1, &0\leq x<1\\2x, &x\geq1\end{cases}$$

\) is continuous. Continuity at the break - points \(x = 0\) and \(x = 1\) was ensured by making the left - hand limit, right - hand limit, and the function value at those points equal.