requesting help on calculus problem

[RayEarth]

verify that the function satisfies the 3 hypotheses of Rolle's theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's theorem.

f(x)=sin2pie*x [-1,1]

Rolle's Theorem states> let f be a function that satisfies the following 3 hypotheses:
1. f is continuous on the closed interval [a,b]
2. f is differentiable on the open interval (a,b)
3. f(a)=f(b)
then there is a number c in (a,b) such that f'(c)=0

I got this far:
1. since it's a trig function, it is continuous
2. sin2pie is differentiable
3. f(-1)=0 and f(1)=0 so f(a)=f(b) therefore there is a number c in (a,b) such that f(c)=0
I differentiate f(x)=sin2pie*x to get: f'(x)=2x*cos2pie + sin2pie
I replace x with c to get -> f'(c)=2c*cos2pie + sin2pie and replace f(c)=0 to get -> 0=2c*cos2pie + sin2pie

the answers in the book say c= 1/4, -1/4, 3/4 and -3/4 I don't see how c can be those numbers and have f'(c)=0


question 2:
show that a polynomial of degree 3 has at most 3 real roots and show that a polynomial of degree n has at most n real roots. I don't even know where to begin on this problem.

 

[JoPh]

question 2:
show that a polynomial of degree 3 has at most 3 real roots and show that a polynomial of degree n has at most n real roots. I don't even know where to begin on this problem.

using what?

 

[JoPh]

for Q2, this is my only idea

draw a graph showing the number of times it could cross the x axis, or show that a x^n can only be divided by x, n times

 

[Mday]

1. rolles theorem states that for any continuous (very pretty very well behaved) function (also differentiable) within an interval, that there is some point not at the endpoints that has the same slope as a linesegment formed by the endpoints of the interval. your interval is from x=-1 to x=1. you have to find what the slope of that line segment is, it's obvious, just graph it.

apparently you cant differentiate.
d/dx (sin(2*pi*x)) = 2*pi*cos(2*pi*x). you have to find all x that satisfies rolles theorem. ie, all x such that f'(x) has the same value as the slope of the linesegment formed by the endpoinst of the interval (x=-1, x=1)

2. it's a proof by contradiction.

suppose that there are more than 3 real roots.
Let's say there are 4, just for arguments sake, then there are 4 numbers, a b c d that are roots of the polynomial. so,
let f(x) be that 3 degree polynomial, since we assumed that there are 4 roots,
f(x)=(x-a)(x-b)(x-c)(x-d)
simplify the function by multiplying out (or something similar)
then you find that the degree of f(x) is not 3 at all, so you have a contradiction.
then you generalize that it is impossible for a polynomial of degree n to have more than n real roots.

 

[RayEarth]

I'm not too sure what you said about my 1st question, I used the product rule to differentiate sin2pie*x by saying
f(x)= sin2pie and g(x)=x therefore I got d/dx[f(x)g(x)]= f(x)d/dx[g(x)] +g(x)d/dx[f(x)]
sin2pie*1 + x*cos2pie*2= sin2pie + 2x*cos2pie


for question 2: so I should say x^3 can only have 3 real root since no polynomial can have more real roots than the highest degree of the x term?

thanks for feedback guys.
this section is about the mean value theorem so I guess I need to use that to solve things here along with using the differentiation skills from the past chapter.