Proof that a function is continuous

In summary, to prove that the function f(x) is continuous when f(x)=0, we can use the fact that the sum of the coefficients of f(x) is zero to find one of the zeros of the function, which is x = 1. Then, we can use long division or synthetic division to find g(x) such that f(x) = (x-1)g(x). We can also take the limit equation and solve for x to find the other zeros, which are -1, 3, and 4. However, simply finding the zeros is not sufficient to show that the function is continuous. We need to show that f(x) is continuous for each of these values of x, which is the
  • #1
LASmith
21
0
Prove that the function is continuous when f(x)=0
f(x)=x4-7x3+11x2+7x-12f(c)-[itex]\epsilon[/itex]<f(x)<f(c)+[itex]\epsilon[/itex]

Limits maybe taken, however, we do not have the value for c in the limit equation.
 
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  • #2
It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

For one of the zeros, notice that the sum of the coefficients of f(x) is zero. Therefore, f(1) = 0.

So you know that one of the factors of f(x) is (x-1). Use long division or synthetic division to find g(x) such that: f(x) = (x-1)g(x).

Added in Edit.
Notice that: f(-x) = x4+7x3+11x2-7x-12. Therefore, f(-(1)) = 0 .
 
Last edited:
  • #3
SammyS said:
It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?
 
  • #4
LASmith said:
Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?
Of course not.

The problem is to show that f(x) is continuous when f(x)=0. So the problem has become: show that f(x) is continuous for x = -1, 1, 3, 4 .
 

1. What is the definition of continuity for a function?

The definition of continuity for a function is that the function has no abrupt changes or breaks in its graph, and it can be drawn without lifting the pencil from the paper. Mathematically, a function is continuous at a point if the limit of the function at that point exists and is equal to the value of the function at that point.

2. How can I prove that a function is continuous?

To prove that a function is continuous, you can use the three-part definition of continuity: the function must have a defined value at the point in question, the limit at that point must exist, and the limit must equal the value of the function at that point. You can also use the epsilon-delta definition of continuity, which involves finding a small enough neighborhood around the point where the function remains within a certain distance of the function value at that point.

3. Can a function be continuous at some points and not at others?

Yes, a function can be continuous at some points and not at others. A function may have abrupt changes or breaks at certain points, making it discontinuous at those points. However, it can still be continuous at other points if the three-part definition of continuity is met at those points.

4. What are some common types of functions that are continuous?

Some common types of functions that are continuous include polynomial functions, rational functions, exponential functions, and trigonometric functions. Additionally, any function that is differentiable at a point is also continuous at that point.

5. Can a function be continuous but not differentiable?

Yes, a function can be continuous but not differentiable. This means that the function meets the three-part definition of continuity at a point, but it does not have a well-defined tangent line or derivative at that point. An example of such a function is the absolute value function at x = 0.

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