Showing That a Function Does Not Have Two Distinct Roots

In summary: Then, according to Rolle's Theorem, there exists some point x=c in the interval 0<a<c<b<1 such that f'(c)=0. Computing f'(c) and finding c we see that c=-1, 1. Since both of these points lie outside the interval (0, 1), they lie outside the interval (a, b). We have reached a contradiction.In summary, there are no roots of f on the interval [0, 1).
  • #1
Expiring
4
3
TL;DR Summary
Showing that the function
x^3 - 3x + m = 0
does not have two distinct roots on the interval 0 <= x <= 1 for any value of m using Rolle's Theorem.
I am wondering if someone can look over my proof, and point out any mistakes I might have made.There is no value of m such that
x^3 - 3x + m = 0
has two distinct roots on the interval 0 <= x <= 1.

Proof.

Let f(x) = x^3 - 3x + m. Suppose, to the contrary, that there is a value of m such that f has two distinct roots in 0 <= x <= 1, and suppose that these roots occur within the interval at x=a and x=b. We apply Rolle's Theorem on the interval (a, b). The conditions of Rolle's Theorem are met since
1.) f is a polynomial, so it is continuous and differentiable everywhere, and, as a result, is continuous and differentiable on (a, b),
2.) f(a)=f(b)=0.
Then, according to Rolle's Theorem, there exists some point x=c in the interval 0<a<c<b<1 such that f'(c)=0. Computing f'(c) and finding c we see that c=-1, 1. Since both of these points lie outside the interval (0, 1), they lie outside the interval (a, b). We have reached a contradiction.
 
Last edited:
  • Like
Likes PeroK
Physics news on Phys.org
  • #2
Expiring said:
TL;DR Summary: Showing that the function
x^3 - 3x + m = 0
does not have two distinct roots for any value of m using Rolle's Theorem.

I am wondering if someone can look over my proof, and point out any mistakes I might have made.There is no value of m such that
x^3 - 3x + m = 0
has two distinct roots on the interval 0 <= x <= 1.
This interval doesn't appear in the original problem description shown in the summary. Is it actually a part of the problem that you neglected to show, or is this something that you added that isn't part of the given problem?
Expiring said:
Proof.

Let f(x) = x^3 - 3x + m. Suppose, to the contrary, that there is a value of m such that f has two distinct roots in 0 <= x <= 1, and suppose that these roots occur within the interval at x=a and x=b. We apply Rolle's Theorem on the interval (a, b). The conditions of Rolle's Theorem are met since
1.) f is a polynomial, so it is continuous and differentiable everywhere, and, as a result, is continuous and differentiable on (a, b),
2.) f(a)=f(b)=0.
Then, according to Rolle's Theorem, there exists some point x=c in the interval 0<a<c<b<1 such that f'(c)=0. Computing f'(c) and finding c we see that c=-1, 1. Since both of these points lie outside the interval (0, 1), they lie outside the interval (a, b). We have reached a contradiction.
 
  • #3
Mark44 said:
This interval doesn't appear in the original problem description shown in the summary. Is it actually a part of the problem that you neglected to show, or is this something that you added that isn't part of the given problem?
It is part of the problem that I mistakenly left out. I fixed my original post.
 
  • #4
With that change, your proof looks fine. You can assume, without loss of generality (wlog), that 0 < a < b < 1.
 

1. How do you show that a function does not have two distinct roots?

To show that a function does not have two distinct roots, you can use the intermediate value theorem. This theorem states that if a function is continuous on a closed interval [a, b] and takes on values of opposite signs at the endpoints, then it must have at least one root between a and b. If the function only has one root in this interval, then it does not have two distinct roots.

2. Can a function have more than two distinct roots?

Yes, a function can have more than two distinct roots. For example, a quadratic function can have two, one, or zero distinct roots, while a cubic function can have three, two, one, or zero distinct roots. The number of distinct roots a function has depends on its degree and the coefficients of the terms.

3. What is the difference between distinct roots and repeated roots?

Distinct roots refer to different values of x that make the function equal to zero. Repeated roots, on the other hand, refer to the same value of x that makes the function equal to zero more than once. In other words, repeated roots are multiple solutions to the same equation, while distinct roots are different solutions to the same equation.

4. Can a function have no distinct roots?

Yes, a function can have no distinct roots. This means that the function does not have any real solutions or values of x that make it equal to zero. For example, a quadratic function with a positive discriminant (b^2-4ac > 0) will have no distinct roots, as the solutions will be complex numbers.

5. How can you prove that a function does not have two distinct roots using calculus?

One way to prove that a function does not have two distinct roots using calculus is by finding the derivative of the function and analyzing its behavior. If the derivative is always positive or always negative, then the function is strictly increasing or decreasing, respectively, and can only have one root. If the derivative changes sign, then the function can have more than one root. Additionally, you can use the second derivative test to determine if the function has a local maximum or minimum, which can help in identifying the number of distinct roots.

Similar threads

I Some questions on l'Hôpital rules
    • Calculus
Replies
9
Views
2K
I Looking for Guarantees that the method of fixed-point iteration will work
    • Calculus
Replies
22
Views
1K
I Why prove Taylor's theorem with p(x)=q(x)−((b−a)^n/(b−x)^n)q(a)?
    • Calculus
Replies
9
Views
858
B Why is this definite integral a single number?
    • Calculus
Replies
20
Views
2K
I Error bounds for Newton's Method
    • Calculus
Replies
13
Views
1K
A Summing over continuum and uncountable numerocities
    • Calculus
Replies
1
Views
941
I Determination of error in interpolating polynomial
    • Calculus
Replies
5
Views
392
I Is My Proof of the Chain Rule Correct?
    • Calculus
Replies
2
Views
1K
I Open interval or Closed interval in defining convex function
    • Calculus
Replies
4
Views
2K
MHB Does the function have a zero in the given interval?
    • Calculus
Replies
9
Views
1K

Hot Threads

Recent Insights

Back
Top