Imvt & LMVT: Existence of Point with f(x)=x

  • Thread starter Ananya0107
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In summary, to show that there exists a point where f(x) = x, we can use the Mean Value Theorem (MVT) and construct a 1 by 1 square on the coordinate plane. By applying MVT, we can find a point c where f'(c) = 1, and then solve for y by finding the value of the integration constant using a point in the domain.
  • #1
Ananya0107
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Let f:[0,1]→[0,1] be a continuous function . Show that there exists a point such that f(x) = x
 
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  • #2
That's a fairly standard homework exercise. What have you tried?
 
  • #3
Tell me if this argument is right... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.
 
  • #4
Apply MVT.
According to MVT, there exists a point c such that, f'(c)={f(1)-f(0)}/(1-0)=1-0=1
Now dy/dx=1. So you can find y.
now substitute a point in the domain to get the value of integration constant.
 

1. What is the definition of IMVT and LMVT?

IMVT stands for Intermediate Value Theorem, which states that if a continuous function f(x) is defined on a closed interval [a, b], then for any value y that lies between f(a) and f(b), there exists at least one value c in the interval [a, b] such that f(c) = y. LMVT stands for Lagrange's Mean Value Theorem, which states that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in the interval (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

2. Why is it important to prove the existence of a point in a function?

Proving the existence of a point in a function is important because it helps us understand the behavior of the function. It also allows us to prove other theorems and solve real-world problems by providing a concrete solution.

3. How do you prove the existence of a point using IMVT?

To prove the existence of a point using IMVT, we first need to show that the function is continuous on the closed interval [a, b]. Then, we need to find two points, a and b, such that f(a) and f(b) have opposite signs. This means that one value is positive and the other is negative. By IMVT, we know that there exists at least one value c in the interval [a, b] such that f(c) = 0. This c is the point we are looking for.

4. Can LMVT be used to prove the existence of a point with any function?

No, LMVT can only be used to prove the existence of a point for functions that are continuous on a closed interval and differentiable on an open interval. Additionally, the function must satisfy the conditions for LMVT, which state that the slope of the tangent line at any point on the function must be equal to the average rate of change of the function over the interval.

5. How is the existence of a point related to the graph of a function?

The existence of a point is related to the graph of a function as it helps us identify specific points on the graph. For example, in the case of IMVT, we know that there exists at least one point on the graph where the function crosses the x-axis. This can help us determine the behavior of the function and make predictions about its values. Similarly, LMVT helps us find points where the slope of the tangent line is equal to the average rate of change of the function, which can also provide valuable information about the function's behavior.

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