Question

A function \(f\) is given on an interval of length \(L .\) In each case sketch the graphs of the even and odd extensions of \(f\) of period \(2 L .\) $$ f(x)=x-3, \quad 0

Short answer

Answer: The even extension of the function in the interval (-4, 4) is: $$ f_e(x) = \begin{cases} x-3, & 0<x<4 \\ -x-3, & -4<x<0 \end{cases} $$ And the odd extension of the function in the interval (-4, 4) is: $$ f_o(x) = \begin{cases} x-3, & 0<x<4 \\ 3-x, & -4<x<0 \end{cases} $$

Step-by-step solution

Even Extension

To find the even extension of \(f(x)\) with period \(8\), we need to make sure the function satisfies \(f(x)=f(-x)\) for all \(x\). First, let's express the current function within the given interval: $$ f(x) = \begin{cases} x-3, & 04} \end{_cases} $$ Now, let's extend it to satisfy the condition \(f(x) = f(-x)\): $$ \text{Even Extension}(x)=f_e(x) = \begin{cases} x-3, & 0<x<4 \\ -x-3, & -4<x<0 \\ \left(\text{repeat for other periods of 8}\right) \end{_cases} $$ To sketch the graph, plot the function \(f(x)=x-3\) for \(0<x<4\) and \(f(x)=-x-3\) for \(-4<x<0\), and keep repeating the resulting graph for every interval of length \(8\).

Odd Extension

To find the odd extension of \(f(x)\) with period \(8\), we need to make sure the function satisfies \(f(x)=-f(-x)\) for all \(x\). First, let's express the current function within the given interval: $$ f(x) = \begin{cases} x-3, & 04 \end{cases} $$ Now, let's extend it to satisfy the condition \(f(x) = -f(-x)\): $$ \text{Odd Extension}(x)=f_o(x) = \begin{cases} x-3, & 0<x<4 \\ 3-x, & -4<x<0 \\ \left(\text{repeat for other periods of 8}\right) \end{cases} $$ To sketch the graph, plot the function \(f(x)=x-3\) for \(0<x<4\) and \(f(x)=3-x\) for \(-4<x<0\), and keep repeating the resulting graph for every interval of length \(8\).