[Maths Class Notes] on Intermediate Value Theorem Pdf for Exam

Intermediate Value Theorem Definition

The intermediate value theorem states that if a continuous function is capable of attaining two values for an equation, then it must also attain all the values that are lying in between these two values. A function is termed continuous when its graph is an unbroken curve.

If we have two points that are connected by a continuous curve, with one point above the line and the other below the line; Intermediate Value theorem says that there will be one place where the curve crosses this line. 

What is Intermediate Value Theorem

For a continuous function f that applies to the interval  [a,b]; the function can take any value between f(a) and f(b) over the interval. For any value of f(a) and f(b); there exists a value of c in [a,b] for which f(c)=L.

Stating in simpler terms, for two real numbers a and b, where a

How is Intermediate Value Theorem Useful as a Property?

Whenever we can show in the equation that there is:

1. A point above the line

2. A point below the line 

3. The curve is continuous; 

we can say that there is a value in between both these points that is true for the equation. 

Intermediate Value Theorem solved Examples and Solutions

If the graph is passing through (a,f(a)) and (b,f(b)), then it must pass through any y-value between f(a) and f(b).

Solved Examples

Example 1:

Is there a value to x5 – 2x3 – 2 = 0 between x=0 and x=2?

When x=0:

05 – 2 × 03 – 2, the value is -2

When x=2:

25 – 2 × 23 – 2, the value is 14

• at x=0, the curve gets below zero

• at x=2, the curve gets above zero

Being a continuous polynomial curve, a curve that contains both variables and coefficients,  it should pass through the curve y=0. Therefore, yes there is a value to the equation x5 – 2x3 – 2 = 0 that exists in the interval [0,2]. 

Example 2:

Write a Formula for the Polynomial Function Shown Below:

This graph has three different x-intercepts: x = –3, 2, and 5. 

Similarly, the y-intercept is exactly located at (0, 2). 

So at both, x = –3 and x = 5, the graph linearly passes through the axis, where it suggests that the corresponding factors of the polynomial function are also linear. But at x = 2, the graph is shown to bounce at an intercept, suggesting that the corresponding factor of the polynomial is quadratic. Putting all these elements together, 

f(x)=a(x+3)(x−2)​2​​(x−5).

To calculate the stretch factor, we can use any other point on the graph as in (0, -2) on the y-intercept to solve the a.

f(0) = a(0+3)(0-2)2(0-5)

-2 = a(0+3)(0-2)2(0-5)

-2 = -60a

a = 1/30

That being said, the graphical polynomial function turns out to be:

f(x)=​1/30(x+3)(x−2)​2​​(x−5).

Example 3:

Through Intermediate Value Theorem, prove that the equation 3x5−4x2=3 is solvable between [0, 2].

Taking m=3,

This given function is known to be continuous for all values of x, as it is a polynomial function. Therefore, it is necessary to note that the graph is not necessary for providing valid proof, but it will help us understand how to use the IVT theorem in this case. In many instances, the problems can often be solved without using the graph. 

f(0)=3(0)5−4(0)2=2<3    

and    

f(2)=3(2)5−4(2)2=80>3

So, f(0)=2

i.e., m=3 is between f(0)and f(2).

As the assumptions of the theorem are now met, we can say that there is a c in the interval [0,2] that satisfies f(c)=m, making the equation solvable.

Proofs of Intermediate Theorem

A standard proof of the Intermediate Value Theorem uses the least upper bound property of the numbers given that every nonempty subset of Real numbers with an upper bound also has a least upper bound. This Intermediate Value property helps in characterizing the real numbers, and one has to remember that the subset of the rational numbers doesn’t have the least upper bound property which is why IVT can’t be applied. Therefore, any proof of the Intermediate Value Theorem appeals to its property equivalent to the upper bound property that uses the whole real numbers.