# 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?

Aug 5, 2016

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.

#### Explanation:

If $f \left(- x\right) = f \left(x\right)$, function is even.

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

$\textcolor{g r e e n}{\text{Function 1}}$

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

$\therefore$ function is even

$\textcolor{g r e e n}{\text{Function 2}}$

$f \left(- x\right) = 4 {\left(- x\right)}^{3} - 2 \left(- x\right) = - 4 {x}^{3} + 2 x = - f \left(x\right)$

$\therefore$ function is odd

$\textcolor{g r e e n}{\text{Function 3}}$

$f \left(- x\right) = 4 {\left(- x\right)}^{2} - 4 \left(- x\right) + 4 = 4 {x}^{2} + 4 x + 4 \ne f \left(x\right) \mathmr{and} - f \left(x\right)$

$\therefore$ function is neither odd nor even

$\textcolor{g r e e n}{\text{Function 4}}$

$f \left(- x\right) = \left(- x\right) - \frac{1}{- x} = - x + \frac{1}{x} = - f \left(x\right)$

$\therefore$ function is odd

$\textcolor{g r e e n}{\text{Function 5}}$

$f \left(- x\right) = \left\mid - x \right\mid - {\left(- x\right)}^{2} + 1 = \left\mid x \right\mid - {x}^{2} + 1 = f \left(x\right)$

$\therefore$ function is even.

$\textcolor{g r e e n}{\text{Function 6}}$

$f \left(- x\right) = \sin \left(- x\right) + 1 = - \sin \left(x\right) + 1 \ne f \left(x\right) \mathmr{and} - f \left(x\right)$

$\therefore$ function is neither even nor odd.