Question #347914

The sum of two discontinuous function is always discontinuous function. True or false with full explanation

1
Expert's answer
2022-06-06T12:29:31-0400

ANSWER . The statement is false.

EXPLANATION

The sum of two discontinuous functions can be a continuous function (example 1), can be a discontinuous function (example 2)

Example 1

h(x)={1 if 2x01 if 0<x2g(x)={x31 if 2x0x3+1 if 0<x2h(x) = \begin{cases}1 & \text{ if } -2\leq x\leq 0 \\ -1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}\\g(x) = \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3}+1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}

These function are discontinuous at x=0x=0 .

f(x)=h(x)+g(x)=x3,x[2,2]f(x)=h(x)+g(x)=x^{3}, x\in[-2,2] is continuous .

Example 2

h(x)={0 if 2x01 if 0<x2g(x)={x31 if 2x0x3+1 if 0<x2h(x) = \begin{cases}0 & \text{ if } -2\leq x\leq 0 \\ -1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}\\g(x) = \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3}+1 & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}

These function are discontinuous at x=0x=0 .

f(x)=h(x)+g(x)={x31 if 2x0x3 if 0<x2f(x) =h(x)+g(x)= \begin{cases}x^{3}-1 & \text{ if } -2\leq x\leq 0 \\ x^{3} & \text{ if } \, \, \, \, \, \, \, 0< x\leq 2 \end{cases}

This function is discontinuous at x=0

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