Is $\frac{(|x| - x^2)}{\sin x}$ an odd function, if it is odd, how do I prove that?

Question

This is the function:

$$f(x)=\frac{|x|-x^{2}}{\sin x}$$

I am not understanding where to go from the second line:

$$\begin{array}{l} f(-x) & =\displaystyle\frac{|-x|-(-x)^{2}}{\sin(-x)}\\ \displaystyle & =\displaystyle\frac{|-x|-(-x)^{2}}{-\sin x} \end{array}$$

How do make this equal $-f(x)$? Or is it neither odd nor even?

$|-x|=|x|,(-x)^2=x^2$, since both are even functions. But $\sin(-x)=-\sin x$ since $\sin x$ is odd. — Shubham Johri, Dec 03 '19 at 13:00

score 0 · Accepted Answer · answered Dec 03 '19 at 12:59

0

We have $|-x|=|x|$ and $(-x)^2=x^2$.

answered Dec 03 '19 at 12:59

Fred

score 0 · Answer 2 · answered Dec 03 '19 at 13:00

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You have $$\begin{array}{l} f(- x) =\frac{|x|\ -\ ( x)^{2}}{-sin( x)}\\ \end{array}$$ and this is $-f(x)$ so $f$ is odd.

answered Dec 03 '19 at 13:00

2 Answers2