Finding Roots and Continuity of Functions

[dekoi]

I recently finished a homework assignment with the exceptions of the following:

1.) f(x) =x^3 - x^2 + x, show there is a number c such that f(c)=10.

f(x) can be equated to 10, but I'm not quite sure how to solve the equation from that point.

2.) Prove that the equation has at least one real root.

e^x = 2 - x

In order to understand this question, i attempted to carry the same procedure with another equation: y = x^2 + x + 2. If the discriminant is 0, then there is a single root. If the discriminant is <0, no roots, and >0, multiple roots. But the same procedure doesn't work with the above equation, or for cubics, quartics, etc.

3.) For what values of x is F continuous?

f(x) = [ 0 if x is rational, 1 if x is irrational

I understand that the function can never be continuous, since it oscillates between 0 and 1 infinitely. But can anyone clarify what the following text means:
http://mathworld.wolfram.com/images/equations/DirichletFunction/equation3.gif

 

[AKG]

1) You want to show that there is some x solving the equation:

x³ - x² + x - 10 = 0

Well you know the polynomial p defined by p(x) = x³ - x² + x - 10 is continuous, so you can apply the intermediate value theorem. That is, if you can find a number x' such that p(x') < 0 and a number x'' such that p(x'') > 0 then you know that there is a number c between x' and x'' (x' may be larger than x'' or may be smaller, doesn't matter) such that p(c) = 0, and such a c will be the c you're looking for.

2) Try same technique as above.

3) You're right that f is not continuous, but are you sure you know why? The function that is 0 on (0,1) then 1 on (1,2) then 0 again on (2,3) then 1 on (3, 4) then 0 on (4,5) then 1 on (5,6), etc. also "oscillates" between 0 and 1, but it has many points of continuity. Also, I don't know if "oscillates" is the correct term. Chances are you know why f is nowhere continuous, and your justification, "since it oscillates between 0 and 1 infinitely" was just an abbreviated version of the actual reasoning, but just in case it's not and you think that statement is sufficient, you should probably do a $\epsilon - \delta$ proof to be sure you know why. Either way, it would be good exercise to do the proof.

What is it that you don't understand about the function $D_M$? It is 0 when x is irrational, and 1/b when x is rational. But what's b? Well when x is rational, it can be expressed as the ratio of two integers, i.e. as a/b. For each rational you know there is a reduced form, so 2/4, 100/200, and 1/2 are all the same number but 1/2 is the reduced form. In this case, b=2, so f(100/200) = f(0.5) = f(1/2) = 1/2. Another example, f(100/150) = 1/3. This function is continuous at every irrational and discontinuous at every rational. Can you prove it?

 

[Werg22]

Here is how to solve a 3rd degree equation;

General cubic form:
y^3 + by^2 + cy + d=0

1.Let y=x-b/3

show that;

x^3 + px + q=0

2.In the equation
x^3 + px + q=0

Let x=H+K, such as 3HK=-p

3.Show that

-q=H^3 + K^3

4.Solve for K in 3HK=-p and replace in H^3 + K^3=-q then show that

H=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3

5. Repeating step 3 and 4 we find out that

K=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3 also.

6.Since x=H+K, then

x=(-q/2 +/- (q^2/4 + p^3/27)^1/2)^1/3 + (-q/2 +/- (q^2/4 +p^3/27)^1/2)^1/3

Then solve for y.

 

[HallsofIvy]

Oh, please don't! The point of this problem was NOT to solve the equation but to use continuity to show that it HAS a solution.
f(x)= x³ - x^2 + x

f(0)= 0, f(1)= 1- 1+1= 2, f(2)= 8- 4+ 2= 6, f(3)= 27- 9+ 3= 21!

I now know, because f(x) is a polynomial and so continuous for all x, that there exist a number x (in fact, between 2 and 3) such that f(x)= 10.

 

[Werg22]

You could also use the odd degree theorem;

x^3 - x^2 + x -10, has to have a solution c such as f(c)=0

Proof: For a large positive value of x^3, x^3> x^2 + 10 - x
then f(x)>0
For a small negative value of x, x^3 <x^2 + 10 - x
then f(x)<0

Since the function is continuous, it has to have a 0.