Lesson Objectives

- Learn about the Intermediate Value Theorem
- Learn about the Upper and Lower Bounds Theorem

## Intermediate Value Theorem, Upper and Lower Bound Rules

In this lesson, we will learn about the Intermediate Value Theorem and the Boundedness Theorem (Upper and Lower Bounds Theorem). These are additional tools that can be used when trying to find the zeros of polynomial functions. The Intermediate Value Theorem will help us to identify intervals where real zeros of polynomial functions are located. The Upper and Lower Bounds Theorem shows how the bottom row of a synthetic division can be used to place upper and lower bounds on possible real zeros of a polynomial function.

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 real zero between -2 and 1. $$f(x)=x^3 - 4x^2 + x + 6$$ We will use the remainder theorem to find f(-2) and f(1). $$f(-2)=-20$$ $$f(1)=4$$ Since f(-2) is negative and f(1) is positive, we know there is at least one real zero between -2 and 1. We can show this using a graph of our polynomial function. Notice the zero occurs at -1. Desmos Link for More Detail

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 synthetic division alternate, we know that -4 is a lower bound. This means there won't be any real zeros that are less than -4.

Example #3: Determine if k is an upper or lower bound. $$f(x)=3x^3 + 4x^2 + 8x + 35$$ $$k=6$$ Since k is positive, we will check to see if k is an upper bound. Since the signs at the bottom of our synthetic division are all non-negative, we know that 6 is an upper bound. This means there won't be any real zeros that are greater than 6.

Example #4: Determine if k is an upper or lower bound. $$f(x)=-x^3 - x^2 + 17x - 15$$ $$k = 4$$ The easiest way to deal with this situation is to work with -f(x). In other words, we are just multiplying through by -1. $$g(x) = -f(x) = x^3 + x^2 - 17x + 15$$ Recall that the zeros will be exactly the same, although the function is different. Compared to f(x), g(x) has been reflected across the x-axis. You can look at the graph below to verify the zeros are exactly the same. Desmos Link for More Detail Returning to the problem, we can just set up our synthetic division. We will use 4 as our value for k, and the coefficients will come from g(x) instead of f(x). Since the signs at the bottom of our synthetic division are all non-negative, we know that 4 is an upper bound. This means there won't be any real zeros that are greater than 4.

In our example, we found the quotient q(x) to be: $$q(x) = 2x^2 - 18x + 88$$ Notice here that the original polynomial has a degree of 3 (odd) and the quotient has a degree of 2 (even). Then as you move to other positions with an even exponent on x, the coefficient will be 0 or positive. $$88 = 88x^0$$ You might recall that x

Since we are plugging in a negative for x and then raising this to an even power, the result will be positive. When this positive is multiplied by a coefficient that is either 0 or positive, the result is non-negative (0 or positive). Recall from our example we used an x-value of -5: $$2(-5)^2 = 2(25) = 50$$ Notice that when we plugged in a negative for x, the even exponent gives us a positive result. (-5)

Then in the positions where we have an odd exponent on x, the coefficient will be either negative or 0. Since we are plugging in a negative for x and then raising this to an odd power, the result will be negative. When this negative is multiplied by a coefficient that is either 0 or negative, the result is non-negative (0 or positive).

Again using our example with an x-value of -5: $$-18(-5)^1 = -18(-5) = 90$$ Notice that when we find the sum, the result will be positive. In our example, we calculated q(x) as: $$2(-5)^2 + (-18)(-5) + 88$$ $$=(2(25) + 90 + 88) = 228$$ So when the n-value is odd and we evaluate q(x) for an x-value that is less than k, we will get a positive number. Putting everything together gives us: $$\text{If} \, x < k{:}$$ $$(x - k)q(x) + r < 0$$ Since we know that (x - k) is negative and q(x) is positive, their product is negative. We also know that r is either 0 or negative. This tells us that (x - k)q(x) + r will be negative and never zero. Therefore f(x) will never be 0 for any x-value less than k. So k is a lower bound when this occurs.

In our example, we found the quotient q(x) to be: $$q(x) = x^3 - 2x^2 + 9x - 58$$ Notice here that the original polynomial has a degree of 4 (even) and the quotient has a degree of 3 (odd). Then as you move to other positions with an odd exponent on x, the coefficient will be 0 or positive. $$9x = 9x^1$$ In our q(x) we have a 1 as the coefficient for x

Since we are plugging in a negative for x and then raising this to an odd power, the result will be negative. When this negative is multiplied by a coefficient that is either 0 or positive, the result is non-positive (0 or negative). Recall from our example we used an x-value of -7: $$1(-7)^3 = (1)(-343) = -343$$ Notice that when we plugged in a negative for x, the odd exponent gives us a negative result. (-7)

Then in the positions where we have an even exponent on x, the coefficient will be either negative or 0. Since we are plugging in a negative for x and then raising this to an even power, the result will be positive. When this positive is multiplied by a coefficient that is either 0 or negative, the result is non-positive (0 or negative).

Again using our example with an x-value of -7: $$-2(-7)^2 = -2(49) = -98$$ $$-58x^0 = -58$$ Notice that when we find the sum, the result will be negative. In our example, we calculated q(x) as: $$(-7)^3 - 2(-7)^2 + 9(-7) - 58$$ $$=-343 -2(49) + 9 (-7) - 58$$ $$=-343 - 98 - 63 - 58 = -562$$ So when the n-value is even and we evaluate q(x) for an x-value that is less than k, we will get a negative number. Putting everything together gives us: $$\text{If} \, x < k{:}$$ $$(x - k)q(x) + r > 0$$ Since we know that (x - k) is negative and q(x) is negative, their product is positive. We also know that r is either 0 or positive. This tells us that (x - k)q(x) + r will be positive and never zero. Therefore f(x) will never be 0 for any x-value less than k. So k is a lower bound when this occurs.

### 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, 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. In the image above, we have a smooth continuous graph of a polynomial function. We can see that f(a) is negative since it lies below the x-axis and f(b) is positive since it lies above the x-axis. Since this curve is continuous, in order to go from below the x-axis to above the x-axis, the graph must cross the x-axis at some x-value of c, where f(c) = 0. Therefore, we will have one real zero between a and b. In other words, if f(a) and f(b) have opposite signs, then 0 is between f(a) and f(b). This tells us that there must be some number c that is between a and b, where f(c) = 0.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 real zero between -2 and 1. $$f(x)=x^3 - 4x^2 + x + 6$$ We will use the remainder theorem to find f(-2) and f(1). $$f(-2)=-20$$ $$f(1)=4$$ Since f(-2) is negative and f(1) is positive, we know there is at least one real zero between -2 and 1. We can show this using a graph of our polynomial function. Notice the zero occurs at -1. Desmos Link for More Detail

### Boundedness Theorem

The Boundedness Theorem, which is also known as the Upper and Lower Bounds Theorem gives us helpful rules for upper and lower bounds for real zeros of polynomial functions. A number is an upper bound if there are no real zeros greater than the number. Similarly, a number is a lower bound if there are no real 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) using synthetic division:- If k > 0 and each number in the bottom row is either positive or zero (non-negative), then k is an upper bound
- f(x) will have no real zero greater than k

- If k < 0 and the numbers in the bottom row alternate in sign, then k is a lower bound
- 0 can count as positive or negative as needed
- f(x) will have no real zero less than k

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 synthetic division alternate, we know that -4 is a lower bound. This means there won't be any real zeros that are less than -4.

Example #3: Determine if k is an upper or lower bound. $$f(x)=3x^3 + 4x^2 + 8x + 35$$ $$k=6$$ Since k is positive, we will check to see if k is an upper bound. Since the signs at the bottom of our synthetic division are all non-negative, we know that 6 is an upper bound. This means there won't be any real zeros that are greater than 6.

### Upper and Lower Bounds Theorem with a Negative Leading Coefficient

Our rule that was given above was written for when our polynomial function has a positive leading coefficient. What happens when the leading coefficient is negative? Let's think about this with an example.Example #4: Determine if k is an upper or lower bound. $$f(x)=-x^3 - x^2 + 17x - 15$$ $$k = 4$$ The easiest way to deal with this situation is to work with -f(x). In other words, we are just multiplying through by -1. $$g(x) = -f(x) = x^3 + x^2 - 17x + 15$$ Recall that the zeros will be exactly the same, although the function is different. Compared to f(x), g(x) has been reflected across the x-axis. You can look at the graph below to verify the zeros are exactly the same. Desmos Link for More Detail

$$f(x) = -x^3 - x^2 + 17x - 15$$

$$g(x) = x^3 + x^2 - 17x + 15$$

### Proof for Upper and Lower Bounds Theorem

We will start with a polynomial function f(x) with real coefficients and a positive leading coefficient. Suppose we divide our polynomial function f(x) by (x - k). We know from the remainder theorem, that we can rewrite our polynomial function as: $$f(x) = (x - k)q(x) + r$$#### Showing that k is an Upper Bound

- All coefficients of q(x), the quotient are non-negative
- The remainder r is non-negative
- k > 0

- x - k > 0
- (x - k) is positive since we are plugging in an x-value that is greater than k

- q(x) > 0
- Evaluating q(x) for an x-value greater than k will give us a positive number
- This comes from the fact that all coefficients of q(x) are non-negative

- r ≥ 0
- The remainder is non-negative (0 or some positive number)
- The remainder doesn't change when we plug in for x here since it is a constant

#### Showing that k is a Lower Bound

- k < 0
- The numbers on the bottom row of the synthetic division alternate in sign
- 0 can be used as positive or negative as needed
- The bottom row will contain n + 1 entries, where n is the degree of the polynomial function

#### Case 1: n, the degree of the polynomial function is odd

We did an example of this earlier so let's revisit that one. $$f(x) = 2x^3 - 10x^2 + 16x - 8$$ $$k = -4$$ Here n is 3, which is an odd number. Let's look at the synthetic division below: We found that -4 was a lower bound. So what happens when we plug in an x-value that is less than -4? Let's rewrite this using the remainder theorem. $$f(x) = (x+4)(2x^2 - 18x + 88) - 360$$ Let's just use -5 but any value less than -4 would also work: $$f(-5) = (-5 + 4)(2(-5)^2 + (-18)(-5) + 88) - 360$$ $$=(-1)(2(25) + 90 + 88) - 360$$ $$=(-1)(228) - 360$$ $$=-228 - 360 = -588$$ We can see that we got a negative value, which will always be the case here. Let's think about why. First off, we know that we are plugging in a negative value for x since k < 0 and we are trying values less than k. What will be true if x < k? $$f(x) = (x - k)q(x) + r$$- x - k < 0
- (x - k) is negative since we are plugging in an x-value that is less than k

- q(x) > 0
- Explained in the paragraph below using our example above

- r ≤ 0
- The remainder is non-positive (0 or some negative number)
- This is true when n is an odd number
- The remainder doesn't change when we plug in for x here since it is a constant

In our example, we found the quotient q(x) to be: $$q(x) = 2x^2 - 18x + 88$$ Notice here that the original polynomial has a degree of 3 (odd) and the quotient has a degree of 2 (even). Then as you move to other positions with an even exponent on x, the coefficient will be 0 or positive. $$88 = 88x^0$$ You might recall that x

^{0}is 1 and 1 times anything is itself. In our q(x) we have a 2 as the coefficient for x^{2}and an 88 as the coefficient for x^{0}.Since we are plugging in a negative for x and then raising this to an even power, the result will be positive. When this positive is multiplied by a coefficient that is either 0 or positive, the result is non-negative (0 or positive). Recall from our example we used an x-value of -5: $$2(-5)^2 = 2(25) = 50$$ Notice that when we plugged in a negative for x, the even exponent gives us a positive result. (-5)

^{2}is 25. That was then multiplied by 2 to give us a result of 50, which is positive. In the case of 88, there isn't anything to plug into so it just stays as 88.Then in the positions where we have an odd exponent on x, the coefficient will be either negative or 0. Since we are plugging in a negative for x and then raising this to an odd power, the result will be negative. When this negative is multiplied by a coefficient that is either 0 or negative, the result is non-negative (0 or positive).

Again using our example with an x-value of -5: $$-18(-5)^1 = -18(-5) = 90$$ Notice that when we find the sum, the result will be positive. In our example, we calculated q(x) as: $$2(-5)^2 + (-18)(-5) + 88$$ $$=(2(25) + 90 + 88) = 228$$ So when the n-value is odd and we evaluate q(x) for an x-value that is less than k, we will get a positive number. Putting everything together gives us: $$\text{If} \, x < k{:}$$ $$(x - k)q(x) + r < 0$$ Since we know that (x - k) is negative and q(x) is positive, their product is negative. We also know that r is either 0 or negative. This tells us that (x - k)q(x) + r will be negative and never zero. Therefore f(x) will never be 0 for any x-value less than k. So k is a lower bound when this occurs.

#### Case 2: n, the degree of the polynomial function is even

$$f(x) = x^4 + 4x^3 - 3x^2 - 4x - 2$$ $$k = -6$$ Here n is 4, which is an even number. Let's look at the synthetic division below: Looking at the results from the synthetic division, we find that -6 is a lower bound. So what happens when we plug in an x-value that is less than -6? Let's rewrite this using the remainder theorem. $$f(x) = (x+6)(x^3 - 2x^2 + 9x - 58) + 346$$ Let's just use -7 but any value less than -6 would also work: $$f(-7) = (-7 + 6)((-7)^3 - 2(-7)^2 + 9(-7) - 58) + 346$$ $$=(-1)(-343 - (2)(49) - 63 - 58) + 346$$ $$=(-1)(-562) + 346 = 562 + 346 = 908$$ We can see that we got a positive value, which will always be the case here. Let's think about why. First off, we know that we are plugging in a negative value for x since k < 0 and we are trying values less than k. What will be true if x < k? $$f(x) = (x - k)q(x) + r$$- x - k < 0
- (x - k) is negative since we are plugging in an x-value that is less than k

- q(x) < 0
- Explained in the paragraph below using our example above

- r ≥ 0
- The remainder is non-negative (0 or some positive number)
- This is true when n is an even number
- The remainder doesn't change when we plug in for x here since it is a constant

In our example, we found the quotient q(x) to be: $$q(x) = x^3 - 2x^2 + 9x - 58$$ Notice here that the original polynomial has a degree of 4 (even) and the quotient has a degree of 3 (odd). Then as you move to other positions with an odd exponent on x, the coefficient will be 0 or positive. $$9x = 9x^1$$ In our q(x) we have a 1 as the coefficient for x

^{3}and a 9 as the coefficient for x^{1}, which is just x.Since we are plugging in a negative for x and then raising this to an odd power, the result will be negative. When this negative is multiplied by a coefficient that is either 0 or positive, the result is non-positive (0 or negative). Recall from our example we used an x-value of -7: $$1(-7)^3 = (1)(-343) = -343$$ Notice that when we plugged in a negative for x, the odd exponent gives us a negative result. (-7)

^{3}is -343. That was then multiplied by 1 to give us a result of -343, which is still negative. In the case of 9x: $$(9)(-7)^1 = (9)(-7) = -63$$ So we see the positive times a negative gives us a negative.Then in the positions where we have an even exponent on x, the coefficient will be either negative or 0. Since we are plugging in a negative for x and then raising this to an even power, the result will be positive. When this positive is multiplied by a coefficient that is either 0 or negative, the result is non-positive (0 or negative).

Again using our example with an x-value of -7: $$-2(-7)^2 = -2(49) = -98$$ $$-58x^0 = -58$$ Notice that when we find the sum, the result will be negative. In our example, we calculated q(x) as: $$(-7)^3 - 2(-7)^2 + 9(-7) - 58$$ $$=-343 -2(49) + 9 (-7) - 58$$ $$=-343 - 98 - 63 - 58 = -562$$ So when the n-value is even and we evaluate q(x) for an x-value that is less than k, we will get a negative number. Putting everything together gives us: $$\text{If} \, x < k{:}$$ $$(x - k)q(x) + r > 0$$ Since we know that (x - k) is negative and q(x) is negative, their product is positive. We also know that r is either 0 or positive. This tells us that (x - k)q(x) + r will be positive and never zero. Therefore f(x) will never be 0 for any x-value less than k. So k is a lower bound when this occurs.

#### 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

Example #3

Determine if k is an upper or lower bound. $$f(x)=x^3 - 2x^2 + 17x + 20$$ $$k=1$$

Please choose the best answer.

A

Upper bound

B

Lower bound

C

Neither

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