Lesson Objectives

- Learn how to use the intermediate value theorem
- Learn how to use the upper bound rule
- Learn how to lower bound rule

## Intermediate Value Theorem, Upper and Lower Bound Rules

In this lesson, we will learn about the intermediate value theorem and the boundedness theorem.

Example #1: Show there is a zero between the two x-values. $$f(x)=x^3 - 4x^2 + x + 6$$ $$-2, 1$$ Let's show there is a zero between -2 and 1. $$f(-2)=(-2)^3 - 4(-2)^2 - 2 + 6$$ $$f(-2)=-8 - 16 - 2 + 6$$ $$f(-2)=-20$$ $$f(1)=(1)^3 - 4(1)^2 + 1 + 6$$ $$f(1)=1 - 4 + 1 + 6$$ $$f(1)=4$$ Since f(-2) is negative and f(1) is positive, we know there is at least one zero between -2 and 1.

Example #2: Determine if k is an upper or lower bound. $$f(x)=2x^3 - 10x^2 + 16x - 8$$ $$k=-4$$ Since k is negative, we will check to see if k is a lower bound. Since the signs at the bottom of our sythetic division alternate, we know that -4 is a lower bound. This means there won't be any zeros that are less than -4.

### The Intermediate Value Theorem

The intermediate value theorem tells us that if f(x) is some polynomial function with only real coefficients, and we have two real numbers, let's say a and b, if the values f(a) and f(b) are opposite in sign, then there exists at least one real zero between a and b. If this is confusing, think about the fact that if we go from positive to negative or from negative to positive, we must cross zero. Therefore, we will have one zero between a and b. A word of caution, the theorem does not say that if f(a) and f(b) are not opposite in sign, that there isn't a real zero between them. It only tells us that if they are opposite in sign, then there will definitely be a real zero between them. Let's look at an example.Example #1: Show there is a zero between the two x-values. $$f(x)=x^3 - 4x^2 + x + 6$$ $$-2, 1$$ Let's show there is a zero between -2 and 1. $$f(-2)=(-2)^3 - 4(-2)^2 - 2 + 6$$ $$f(-2)=-8 - 16 - 2 + 6$$ $$f(-2)=-20$$ $$f(1)=(1)^3 - 4(1)^2 + 1 + 6$$ $$f(1)=1 - 4 + 1 + 6$$ $$f(1)=4$$ Since f(-2) is negative and f(1) is positive, we know there is at least one zero between -2 and 1.

### Boundedness Theorem

The boundedness theorem gives us helpful rules for upper and lower bounds for zeros of polynomial functions. A number is an upper bound if there are no zeros greater than the number. Similarly, a number is a lower bound if there are no zeros less than the number. Suppose we have a polynomial function f(x) with real coefficients and a positive leading coefficient. If we divide f(x) by (x - k):- If k > 0 and each number in the last row is either positive or zero, then k is an upper bound
- If k < 0 and each number in the last row are alternating in signs then k is a lower bound
- 0 can count as positive or negative as needed

Example #2: Determine if k is an upper or lower bound. $$f(x)=2x^3 - 10x^2 + 16x - 8$$ $$k=-4$$ Since k is negative, we will check to see if k is a lower bound. Since the signs at the bottom of our sythetic division alternate, we know that -4 is a lower bound. This means there won't be any zeros that are less than -4.

#### Skills Check:

Example #1

Determine if k is an upper or lower bound. $$f(x)=3x^4 - 6x^3 - 6x + 4$$ $$k=-1$$

Please choose the best answer.

A

Upper bound

B

Lower bound

C

Neither

Example #2

Determine if k is an upper or lower bound. $$f(x)=x^3 - 6x^2 + 10x - 8$$ $$k=6$$

Please choose the best answer.

A

Upper bound

B

Lower bound

C

Neither

Congrats, Your Score is 100%

Better Luck Next Time, Your Score is %

Try again?

Ready for more?

Watch the Step by Step Video Lesson Take the Practice Test