Is the function h(x) odd, even, or neither?

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Is the function h(x) odd, even, or neither?

This is the problem I'm stuggling on. $h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$

even formula y=f(-x)
odd formula y=-f(x)

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Answer 1 · 2017-10-04T06:31:51

It is an odd function. See explanation.

Explanation:

To check if the function $h (x)$ $h(x)$ is odd, even, or neither you have to calculate the value of $h (- x)$ $h(-x)$ and compare it to the original function.

Here we have:

$h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$

$h (- x) = \frac{2 {(- x)}^{2}}{5 {(- x)}^{3}} = \frac{2 x^{2}}{- 5 x^{3}} = - \frac{2 x^{2}}{5 x^{3}} = - h (x)$ $h(-x)=(2(-x)^2)/(5(-x)^3)=(2x^2)/(-5x^3)=-(2x^2)/(5x^3)=-h(x)$

We calculated, that $h (- x) = - h (x)$ $h(-x)=-h(x)$ , this is the condition of the odd function, so $h (x)$ $h(x)$ is odd.

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Is the function h(x) odd, even, or neither?

This is the problem I'm stuggling on. $h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$

even formula y=f(-x)
odd formula y=-f(x)

1 Answer

Explanation:

$h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$

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Math

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Is the function h(x) odd, even, or neither?

This is the problem I'm stuggling on.h(x)=2x25x3h(x)=(2x^2)/(5x^3) even formula y=f(-x) odd formula y=-f(x)

1 Answer

Explanation:

h(x)=2x25x3h(x)=(2x^2)/(5x^3)

Related questions

Impact of this question

This is the problem I'm stuggling on. $h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$

even formula y=f(-x)
odd formula y=-f(x)

$h (x) = \frac{2 x^{2}}{5 x^{3}}$ $h(x)=(2x^2)/(5x^3)$