Questions
Find $a,b,c,d$ so the function $f$ is an even function.
Find $a,b,c,d$ so the function $f$ is an odd function.
My understanding:
Even function means $f(x) = f(-x)$.
Odd function means $-f(x) = f(-x)$
We will have 2 equations, one for even function and the other for odd function:
$ax^3+bx^2+cx+d = -ax^3+bx^2-cx+d$
$-(ax^3+bx^2+cx+d) = -ax^3+bx^2-cx+d$
If I cancel out what I can I'll have:
$ax^3 + cx = -ax^3 - cx$ $<=>$ $2(ax^3+cx) = 0$ $<=>$ $ax^3+cx = 0$ $<=>$ $x*(ax^2+c)=0$
$-bx^2-d = bx^2 + d$ $<=>$ $bx^2+d = 0$
Should I not cancel? Any hints to how I should continue? First equation is basically $x = 0$ or $ax^2+c = 0$
Indeed, your reasoning is correct so far. In the even case, since $b$ and $d$ canceled out (thus they can actually be anything), you are only left with the equation you mentioned, $x(a x^2 + c) = 0$ for all $x \in \mathbb{R}.$ But this is obviously not true if $a$ and $c$ are not $0$, so we should in fact have $a = c = 0.$ Thus, the general solution for the first part is $f(x) = b x^2 + d.$ Doing the same for the odd case, we get the general solution $f(x) = a x^3 + cx.$