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

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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 $f (- x) = f (x)$ $f(-x) = f(x)$ , function is even.

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

$Function 1$ $color(green)("Function 1")$

$f (- x) = 2 {(- x)}^{2} + 7 = 2 x^{2} + 7 = f (x)$ $f(-x) = 2(-x)^2 + 7 = 2x^2 + 7 = f(x)$

$therefore$ 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.