These questions give me a problem since the rules of distribution seem not to apply to them, for example:
Determine whether the following function is even, odd, or neither? $g(x)= 1-x^4 $
$g(-x) =1 - (-1x)^4$
$1-(+1x^4)$
distribute the negative $1-1x^4$
factor out the negative
$-1 (-1 +1x^4)$
Thus this function is neither, however, the book says it is even, why?
Why are the rules of distribution of the negative sign not being applied here?
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$\begingroup$Recall that $(-x)^2 = (-x)(-x) = x^2\geq 0 \forall x$. And note that $(-x)^4 = ((-x)^2)^2 = (x^2)^2 = x^4$.
Indeed, $(-x)^{2n} = x^{2n}$ when $n \in \mathbb Z$.
$$g(-x) = (1-(-x)^4) = 1 - x^4 = g(x) $$
which is what you found after "distributing the negative": $1 - 1\cdot x^4 = 1- x^4$.
So, what does it mean when $g(x) = g(-x)?$
$\endgroup$ 3 $\begingroup$The definition of even function is $\forall x$ $$f(x)=f(-x)$$ The definition of odd function is $$f(x)=-f(-x)$$ The function is neither iff $$f(x)\not=\pm f(-x)$$ From those definitions can you conclude why the function is even?
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