How do determine whether these relations are even, odd, or neither: #f(x)=2x^2+7#? #f(x)=4x^3-2x#? #f(x)=4x^2-4x+4#? #f(x)=x-(1/x)#? #f(x)=|x|-x^2+1#? #f(x)=sin(x)+1#?

Question

6156 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-05T17:13:58

Function 1 is even.
Function 2 is odd.
Function 3 is neither.
Function 4 is odd.
Function 5 is even.
Function 6 is neither.

Next time, try and ask separate questions rather than lots of the same at once, people are here to help you, not to do your homework for you.

If #f(-x) = f(x)#, function is even.

If #f(-x) = -f(x)#, function is odd.

#color(green)("Function 1")#

#f(-x) = 2(-x)^2 + 7 = 2x^2 + 7 = f(x)#

#therefore# function is even

#color(green)("Function 2")#

#f(-x) = 4(-x)^3 - 2(-x) = -4x^3 + 2x = -f(x)#

#therefore# function is odd

#color(green)("Function 3")#

#f(-x) = 4(-x)^2 - 4(-x) + 4 = 4x^2 + 4x + 4 != f(x) or -f(x)#

#therefore# function is neither odd nor even

#color(green)("Function 4")#

#f(-x) = (-x) - (1)/(-x) = -x + 1/x = -f(x)#

#therefore# function is odd

#color(green)("Function 5")#

#f(-x) = abs(-x) - (-x)^2 + 1 = abs(x) - x^2 + 1 = f(x)#

#therefore# function is even.

#color(green)("Function 6")#

#f(-x) = sin(-x) + 1 = -sin(x) + 1 != f(x) or -f(x)#

#therefore# function is neither even nor odd.