Limits, Rational roots, and inequalities

[mmm4444bot]

Why did you choose to calculate P(1), to start?

You were doing it to see if x = 1 is a zero of P(x), right?

But why did you choose to start experimenting with x = 1?

1 and 0 are both good numbers to start with, if you're experimenting.

But we don't have to guess.

The Rational Roots Theorem provides us with a list of ALL possible Rational roots of the polynomial.

No, guessing and testing possible zeros is not the theorem. The theorem gives us the numbers to try.

 

[funnytim]

Oh...in math last year, I was taught to 'guess and check'. I know there's a way to find out the number (Rational Root Theorem) but I think I'll review it another day.
Right now tryign to tackle the graphing question...

 

[funnytim]

Question:

Solve f '(x) = 0, to find them.

x = ±sqrt(2)/2 (i.e., halfway between the intercepts and the origin)

When solved for f '(x), this is what I did:
f ' (x) = -4x^3 +2x
2x (-2x^2 + 1) = 0

x=0 , AND
2x^2 = 1
x = +/- (square root 1/2)

How come you got something x = ±sqrt(2)/2 ? What did I do wrong? Thx yet again

 

[mmm4444bot]

When solved for [f '(x) = 0], this is what I did:

2x (-2x^2 + 1) = 0

…

x = +/- (square root 1/2)

How come you got something x = ±sqrt(2)/2 ?

Your results are okay.

I rationalized the denominator.

sqrt(1/2) = sqrt(2)/2

The rationalized-denominator form is nice because it clearly shows that the x-coordinates of the local maxima (excluding the y-intercept) are halfway between the x-intercepts and the origin.

In other words, no calculator needed to estimate the square root of 1/2.

\(\displaystyle \sqrt{\frac{1}{2}} \;=\; \frac{\sqrt{1}}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}} \;=\; \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \;=\; \frac{\sqrt{2}}{2}\)

 

[funnytim]

Yea I know, basic stuff lol.

OK, i think I've finished most of my assignment more or less.

Thank you very very very much for your help, I am very very grateful for it. This is a great forum and I'll definitely be back for more help .

Thanks again!

(And now it's time to sleep...finally!!! Also had to travel back to my campus today, so kinda tiring).

 

[mmm4444bot]

[attachment=0:1p8rlco2]Humps.JPG[/attachment:1p8rlco2]

 

[funnytim]

Yup, mine is something like that too (or at least, I tried to)

Thanks again!