Question

Can you please explain to me which functions are odd?

Answer 1

A function is odd if f(-x)=-f(x).
So given a function, to check whether it is odd or not, we calculate f(-x). If it is equal to -f(x), then the function is odd; if not, it is not odd.


$\displaystyle{f(x)= -\frac{1}{2}x^4+5$

$\displaystyle{f(-x)= -\frac{1}{2}(-x)^4+5= -\frac{1}{2}x^4+5\neq-f(x)$

thus the function is not odd


$\displaystyle{f(x)=-8x^3+5x$

$\displaystyle{f(-x)=-8(-x)^3+5(-x)=8x^3-5x=-(-8x^3+5x)=-f(x)$

thus the function is odd



$\displaystyle{f(x)=- \frac{4}{x^3}-x+1$

$\displaystyle{f(-x)=- \frac{4}{(-x)^3}-(-x)+1= \frac{4}{x^3}+x+1\neq-f(x)$

thus the function is not odd



$\displaystyle{f(x)= \frac{x^5}{x^4-1}$

$\displaystyle{f(-x)= \frac{(-x)^5}{(-x)^4-1}= \frac{-x^5}{x^4-1}=- \frac{x^5}{x^4-1}=-f(x)$

thus the function is odd



$\displaystyle{f(x)=-\sqrt{2x}$

$\displaystyle{f(-x)=-\sqrt{2(-x)}= -\sqrt{-2x}$

In this particular case f(x) and f(-x) can both exist only for x=0 (because one of them is certainly negative). Thus the function is not odd


$\displaystyle{f(x)=3 \sqrt{x} -x^3$

similarly to the previous case,  the Domain of f is [0, infinity) and f(-x) cannot be calculated except for x=0. So the function is not odd.