# Is the function f(x) = x^6-2x^2+3 even, odd or neither?

Nov 4, 2015

$f \left(x\right)$ can be either of odd or even.

#### Explanation:

Let n be any even number
Let m be any other even number
(Note m could take on the same value as n)

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Then if $x$ is odd we can write $x \to n + 1$
Then ${\left(n + 1\right)}^{2} \to {n}^{2} + 2 n + 1$
Then ${\left(n + 1\right)}^{3} \to {n}^{3} + 3 {n}^{2} + 3 n + 1$
There will always be a $1 \times 1$ which when added to all the even numbers (n's) makes the total odd.

If $x$ is divisible by 2 then it is even and ${x}^{6}$ will also be even.

$\textcolor{b l u e}{{x}^{6} \text{ can be either even or odd}}$
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If ${x}^{2}$ is odd then let it be represented by $\left(n + 1\right)$
We have $2 {x}^{2}$ which is $\left(n + 1\right) + \left(n + 1\right)$ which becomes $2 n + 2$ which is even.

$\textcolor{b l u e}{2 {x}^{2} \text{ is always even}}$
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$\textcolor{b l u e}{3 \text{ is always odd}}$

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So it boils down to the behaviour of combining 3 and ${x}^{6}$

Let ${x}^{6} \to \left(n + 1\right) \to \text{odd}$
Let $3 \to \left(m + 1\right) \to \text{odd}$

$\textcolor{red}{\text{Then "x^6 + 3 -> (n+1) + (m+1) = m+n+2" which is even}}$

Likewise if n is even $\textcolor{red}{\text{then "x^6 + 3 -> (n) + (m+1) = m+n+1" which is odd}}$
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Conclusion:
$f \left(x\right)$ can be either of odd or even.