Question

Math 180/Pre-Calculus; Rational Zeros of Polynomials.

Use the function P(x)=x^3-5 (LaTeX below) along with the IVT and the process presented here to determine an estimate for the cube root of 5 correct to ONE decimal place, and then correct to TWO decimal places of accuracy. Show all work and each step. Explain your results along the way.

(IVT/Intermediate Value Theorem stated below; work so far will also be shown)

Answer 1

LaTeX equation:\[P(x)=x^{3}-5\]Intermediate Value Theorem: states - "If P is a polynomial and if P(a) and P(b) are of opposite sign, then P has a zero between a and b."
ALSO STATED - The Intermediate Value Theorem is an example of an existence theorem - it tells us that a zero exists, but doesn't tell us exactly where it is. Nevertheless, we can use the theorem to zero in on the zero.

Answer 2

So far I have gotten this:Equation\[x^{3}-5\]Rough value range\[P(4)\rightarrow P(5);\]\[P(4)=4^{3}-5=64-5=59\rightarrow P(5)=5^{3}-5=125-5=120\]This is where I get lost; what I've learned from applying the theorem is that the range between these two values should have a negative minimum and a positive maximum. :\

Answer 3

aka P(4) would yield a negative answer and P(5) a positive... but of course this is not the case. Have I done something wrong or am I on the right path?

p.s. I used the x-values 4 and 5 because of the sqrt(5).

Answer 4

Cube root of 5 is not between 4 and 5. It has to be between 1 and 2 since 1^3=1 and 2^3 =8.

Answer 5

oh right. I thought it was square root D:

Answer 6

My bad, sorry about that.

Answer 7

Square root of 5 couldn't be between 4 and 5 either, though... 4^2 is 16

Answer 8

But the example problem placed sqrt(2) between 1 and 2, I don't know why...\[\sqrt{2}\rightarrow P(1),P(2)\]The example says, and I quote, "Evaluate P(1) and P(2). State why the intermediate value theorem tells us that a zero must exist between x=1 and x=2"

Answer 9

I understand what you were saying, though. I am just confused.

Answer 10

Because the square root of 2 is between 1 and 2. 1^1 =1 and 2^2 = 4

Answer 11

Oh! Okay, I see :-o
Thank you, I will come back to this thread if I have anymore questions then :-)