Even & Odd Functions

[sinful sparrow]

I do not understand how to graph this function and create a rule for it, can someone please explain?

[sinful sparrow]

<@&286206848099549185>

[old spire]

as u know, it is gonna be a parabola for x>0. As odd functions have rotational symmetry, this would be looking like -x^2 for x<=0. Because f(-x) has to equal -f(x). This will only happen if for x <=0 the function is off opposite gradient and values of the function's behaviour at x >0

the rule for this function would be thru a piecewise function

f(x) : { x^2, x>0
-x^2, <= 0

[sinful sparrow]

thats where im having trouble, im supposed to format it like this:

[old spire]

hmm ya doesnt look like a piecewise can fit in there

what i think might be the case is

they might have restricted the domain of f(x) for x >0 (i think and hope)

but that would make it too easy

i hate how they didnt specified a domain

[sinful sparrow]

yes its a really annoying question

i understand the concept but the question is trippy

[old spire]

which means we had to do it ourselves thru a piecewise, and now they aint accepting that??? all that says to mke is to give them a taste of their own medicine --> just say x^2 really

[sinful sparrow]

i already tried that

[old spire]

gotcha

[old spire]

it doesnt accept?

[sinful sparrow]

it doesnt

[faint ivy]

Why not like x^2 * x/|x|

[old spire]

try this @sinful sparrow

but

[faint ivy]

Or just x * |x|

[sinful sparrow]

ya so x^3/|x|

[old spire]

ahh nice this should work

[sinful sparrow]

it worked

[old spire]

so the trick was we dont have to split the mod

[sinful sparrow]

x^3/|x| worked

[old spire]

LESGOO

@faint ivy @sinful sparrow gg

[sinful sparrow]

btw i had tried x^3/x earlier and i was surprised that it didn't work but now i see why

[old spire]

ye

[sinful sparrow]

ty guys

[old spire]

np, a good way to learn for me asw

[sinful sparrow]

if you have time, can you help me with this one as well?

its the other one i was stuck on

[old spire]

yeah sure

this one u have to use long division, then u will get a fraction and a constant added

*function of x and constant added

first step - divide 2x^3 + x^2 + 1 by x^2 - 9

[faint ivy]

The even function is 1+x^2 / x^2 + 9

The odd function is 2x^3 / x^2 + 9

[sinful sparrow]

already tried that i think

let me try again

[faint ivy]

its definitely right

[sinful sparrow]

wait... x^2+9 for the denominator?

[faint ivy]

Oops

-9

For both

Read it wrong

It’s still the correct even / oddness

Tho

[sinful sparrow]

it says its wrong

i already tried it

[old spire]

does it take 2x+1 for the odd part?

[sinful sparrow]

thats what i did

(highlighted boxes are incorrect)

[sinful sparrow]

i only have 2 attempts left so i want to do even with odd

[old spire]

yeh i was gonna give the even part if this one worked

[sinful sparrow]

wait i think i know what i did wrong

g(x) is supposed to be odd

[old spire]

but i think 2x+1 and -18x+10/x^2-9 are both odd

[sinful sparrow]

oops

[old spire]

bruh

Lol

[sinful sparrow]

lmao

its correct now lol

my bad

[old spire]

this was thru long division btw, so if the question asked for a sum of odd functions youd go this way

[sinful sparrow]

oh I havent learned that method of even & odd functions yet

we just started the lesson

[old spire]

nah its not really a method, just using past knowledge.

[sinful sparrow]

oh ok

[old spire]

but those r good questions

a hyperbola will always be odd

and 2x+1 is also odd

its good application

[sinful sparrow]

ya these kinds of questions are easy but not all of them are like this

the question just asks if the function is even or odd

ok cya

i might have more questions later tho

[old spire]

yeh sure

[sinful sparrow]

.solved

[frail galleon]

Solved

Post marked as solved by @sinful sparrow.

Use .unsolved if this was a mistake.

[sinful sparrow]

@old spire isn't this correct?

.unsolved

[frail galleon]

Unsolved

Post marked as unsolved by @sinful sparrow.

Use .solved to mark as solved.

[sinful sparrow]

i used discriminant but it says its wrong

[faint ivy]

The fundamental theorem of algebra states that any polynomial with degree n will have exactly n roots, so one way to lower the roots of this polynomial to 1 is by lowering its degree to 1. This can be achieved by setting n to zero, which removes the first term.

[sinful sparrow]

oh ur right i hadn't thought of that

[old spire]

discriminant should work

[sinful sparrow]

i somehow forgot i could set n to 0

lol

.solved

[frail galleon]

Solved

Post marked as solved by @sinful sparrow.

Use .unsolved if this was a mistake.

[faint ivy]

the discriminant is 4 + n16

Setting n to -1/4 does give a polynomial with "one" root

but its technically zero twice

in the image it also looks like you typed a space after the number 1

which could be causing an issue

[sinful sparrow]

no thats just the formatting

the error was that i didnt put n=0 as a solution

[faint ivy]

thats strange

as both cause it to have one root

[sinful sparrow]

no it always asks for all solutions

[faint ivy]

ah

so you need 0, -1/4

[sinful sparrow]

altho the way the question is worded makes it sound like n has only one value

[faint ivy]

yeah

because it says "the value" instead of the values

implying that there is only one

[sinful sparrow]

@faint ivy is this not correct?

[faint ivy]

?

[sinful sparrow]

its another question

my answer is in the box

[faint ivy]

that does not immediately appear to be correct

the 2 in the (x^2) should add some extra x terms

but you still have -2

[sinful sparrow]

i just did 5(x)^2-2x+3+5(2)^2-2(2)+3

[faint ivy]

5(x + 2)^2 = 5(x + 2)(x + 2) = 5(x^2 + 4x + 4)

[sinful sparrow]

oh wait ya

i forgot i have to substitute them together

f(x+y) != f(x)+f(y)

[faint ivy]

mhm

unless f is a linear transformation

[sinful sparrow]

so its 5x^2+18x+19