Limits, Rational roots, and inequalities



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.

 
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...
 
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
 
funnytim said:
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.

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

 
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).
 
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