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No joke, I got a prob on a quiz for my AP BC calc class and I miserably got it wrong...

the lesson mentioned inequalities for polynomials, but not polynomial abs value probs.

And, if I'm supposed to use a graphing calculator, tell me, because I honestly have no idea.

Yes, but linear absolute value problems can have extraneous solutions...

then check the possible answers in the original equation

So I probs just want a graphing calculator?

off hand, I don't know. do you have a specific example?

Let me access my course. Sec

|x^3-x^2+x-1|<3
"Round your answer to three decimal places and place in ascending order with a comma in between."
Answer: -0.811

Alright. Thanks for looking at it with me.

if you graph the inner polynomial, whenever it goes negative, it just mirrors above the x axis

well, the curve does, then pick a point that is clearly not on the curve to test

|dw:1348175550151:dw|

|dw:1348175643801:dw|

o.o

it's a fake 6 degree polynomial!

|dw:1348176111960:dw|

Mmm.

im curious why the y=3 line

i see it now

yes, that y=3 line represents where the x axis would be if we subtract 3 from each side

|x^3-x^2+x-1| < 3
consider the values such that
(x^3-x^2+x-1) = 3
-(x^3-x^2+x-1) = 3

x^3-x^2+x-4 = 0
and
x^3-x^2+x+2 = 0
there is no good way to do a general cubic solution is there

yeah I'm not cardano, I have 30 minutes to do this xD. I think it was a calculator prblem lol

do you need exact or approx ?

approx.

x = 1.743 and -0.811 is what the google gives me as approx zeros

we can test those in the original format

kay

I see. So that's what I'm supposed to do. Thank you.

Sounds like you are ready for these... give 'em hell

It's not even calculus and I got it wrong. xS