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how would you find algebracically if this is an odd or even function± 1-^3 square root (x)(little 3; squareroot symbol; x underneith the radical)

iris014

by iris014 at November 03, 2009

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I'm not sure what the 1- is (what's being cubed). But to find out if it's odd or even, plug in a negative x where x is. Since I'm assuming you're not working with complex numbers, either you have an absolute value around the x, in which case you would get exactly the same function (even, or odd if 1- is x), or you have a restricted domain (only on the right side of the y axis) and it's neither odd nor even because you can't put a negative x in.

Alison037 Alison037 November 03, 2009

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Just to show the work algebriacally, if it's x^3 sqrt(|x|) (notice x is absolute valued) f(x) = x^3 sqrt(|x|) --> f(-x) = (-x)^3*sqrt(|-x|) = -1*x^3*sqrt(|x|) = -(x^3*sqrt(|x|)) = -f(x) since f(-x) = -f(x), the function is odd. If the 1- was not x or something with x, it wouldn't be affected and the negative one would never be in the front, so you would have f(-x) = f(x), or an even function.

Alison037 Alison037 November 03, 2009

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