Question

Determine whether the piecewise function is odd or even ??

Answer 1

whats the function?

Answer 2

I know that f(−x)=f(x) is even and f(−x)=−f(x) is odd. However, I can't see a way to apply this rule here to find a solution.

Answer 3

\[\LARGE f(t) = \begin{cases}e^{t} \ , \ -\pi \le t < 0\\ {}\\e^{-t} \ , \ 0 \le t < \pi\\\end{cases}\]

Answer 4

Replace every `t` with `-t`. Then simplify.

\[\LARGE f(t) = \begin{cases}e^{t} \ , \ -\pi \le t < 0\\ {}\\e^{-t} \ , \ 0 \le t < \pi\\\end{cases}\]


\[\LARGE f({\color{red}{t}}) = \begin{cases}e^{{\color{red}{t}}} \ , \ -\pi \le {\color{red}{t}} < 0\\ {}\\e^{-{\color{red}{t}}} \ , \ 0 \le {\color{red}{t}} < \pi\\\end{cases}\]


\[\LARGE f({\color{red}{-t}}) = \begin{cases}e^{{\color{red}{-t}}} \ , \ -\pi \le {\color{red}{-t}} < 0\\ {}\\e^{-({\color{red}{-t}})} \ , \ 0 \le {\color{red}{-t}} < \pi\\\end{cases}\]



\[\LARGE f({\color{black}{-t}}) = \begin{cases}e^{{\color{black}{-t}}} \ , \ \pi \ge {\color{black}{t}} > 0\\ {}\\e^{t} \ , \ 0 \ge {\color{black}{t}} > -\pi\\\end{cases}\]



\[\LARGE f({\color{black}{-t}}) = \begin{cases}e^{{\color{black}{-t}}} \ , \ 0 < {\color{black}{t}} \le \pi\\ {}\\e^{t} \ , \ -\pi < {\color{black}{t}} \le 0\\\end{cases}\]

Answer 5

What do you notice about `f(t)` and `f(-t)` ?

Answer 6

both functions have interchanged their position

Answer 7

They're effectively the same

\[\LARGE f(t) = \begin{cases}e^{t} \ , \ -\pi \le t < 0\\ {}\\e^{-t} \ , \ 0 \le t < \pi\\\end{cases}\]


\[\LARGE f({\color{black}{-t}}) = \begin{cases}e^{{\color{black}{-t}}} \ , \ 0 < {\color{black}{t}} \le \pi\\ {}\\e^{t} \ , \ -\pi < {\color{black}{t}} \le 0\\\end{cases}\]

Answer 8

the only difference really is the endpoints are slightly different

Answer 9

so this means the function is even..right ???

Answer 10

yes if we don't worry about the endpoints, then `f(t) = f(-t)` is true for every t value in the domain

Answer 11

ok thank you very much

Answer 12

you're welcome