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How do you determine if #y=2x^5+x# is an even or odd function?

Answer 1

Benjamin Achtenberg

#y=2x^5+x# is an odd function

If #y_((-x)) = y_x# then the function is even

If #y_((-x))=-y_x# then the function is odd

Given #color(white)("XXX")y_x=2x^5+x#

then #color(white)("XXX")y_((-x)) = 2(-x)^5+(-x)#

#color(white)("XXXXXXX")=-2x^5-x#

#color(white)("XXXXXXX")=-(2x^5+x)#

#color(white)("XXXXXXX")=-y_x#

So the function is odd

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Answer 2

Theodore Amidon

To determine if the function y = 2x^5 + x is even or odd, we can examine its symmetry properties.

A function is even if f(-x) = f(x) for all x in its domain. A function is odd if f(-x) = -f(x) for all x in its domain.

To check for evenness, we substitute -x into the function: f(-x) = 2(-x)^5 + (-x) = 2(-x^5) - x = -2x^5 - x

To check for oddness, we substitute -x into the function: f(-x) = 2(-x)^5 + (-x) = 2(-x^5) - x = -2x^5 - x

Since f(-x) = f(x) for this function, it is even.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.