Determining the Existence of any Intersection Between a

Logarithmic Function and a Polynomial

Yizhou (Ethan) Xu

Mathematics

Supervisor: Ms. Kobbi

Sir Winston Churchill Secondary School

May 2016

Word Count: 3975

Acknowledgements: Special thanks to Ms. Kobbi and Dr. Mark

Krusemeyer who has provided helpful insights throughout this research

Abstract

This research paper aims to find mathematical formulae that can be used to predict the

existence of intersections between a logarithmic function and a polynomial. To narrow the focus

of this paper, several assumptions have been made: intersections between the two functions at

infinity are not considered as intersections, all coefficients of the two functions are real numbers

other than ± ∞, and the base of the logarithmic function is greater than 1. Case analysis was used

as the main approach. The degrees of the two functions were varied in each case, and their

behaviour was analyzed. For each case, I established an inequality between the coefficients of the

two functions for determining if there are intersections between the functions.

Word Count: 119

Table of Contents

Introduction------------------------------------------------------------------------------------1

1. deg (f) = 0 or deg (g) = 0------------------------------------------------------------------3

2. deg (f) = 1, deg (g) = 1---------------------------------------------------------------------3

3. deg (f) = 2, deg (g) = 1---------------------------------------------------------------------6

4. deg (f) = 1, deg (g) = 2--------------------------------------------------------------------10

5. deg (f) = 2, deg (g) = 2-------------------------------------------------------------------12

6. Further observation-----------------------------------------------------------------------21

7. Conclusion---------------------------------------------------------------------------------22

Bibliography----------------------------------------------------------------------------------24

1

Introduction

My research question asks: are there mathematical formulae to predict whether the

equation, log𝑎 𝑓(𝑥) = 𝑔(𝑥), where 𝑓(𝑥) and 𝑔(𝑥) are polynomials, has any solutions.

The reason why I chose to investigate this question is that, from my past algebraic problem-

solving experiences, I often found it difficult to determine whether a solution exists for a given

equation. In order to solve this problem, I strive to find a quicker and more convenient way for

solving these questions, which can vastly increase the efficiency.

In my research paper, I have limited the scope of my investigation to a specific type of

equation: one between a logarithmic function and a polynomial function. There may be ways to

determine if intersections exist for other types of equations.

My research integrates three fields of mathematics: calculus, algebra, and geometry. Calculus

will be mainly employed to calculate the slopes of log𝑎 𝑓(𝑥) and 𝑔(𝑥) at a given point in order

to determine when the two functions do not intersect. Finding the slopes of log𝑎 𝑓(𝑥) and 𝑔(𝑥)

will be very helpful for determining the x-value at which these functions lie tangent to each other.

I will use algebra to manipulate the two functions and their derivatives to yield a restriction on

the coefficients of 𝑓(𝑥) and 𝑔(𝑥) such that there exists no intersections between them.

Geometry will be mainly used to predict the behaviour of log𝑎 𝑓(𝑥) and 𝑔(𝑥) in order to

determine how the restrictions on the coefficients of these functions should be placed such that

they do not intersect.

2

I will first limit the degrees of the functions 𝑓(𝑥) and 𝑔(𝑥) to 1 in order to gain a better

understanding of my investigation. Afterwards, I will vary the degrees of these functions to

generalize my results so that they can be applied to more scenarios.

The solution to my research question will provide a relatively simple way to solve these

equations. Its significance lies in the fact that it is especially useful for those who have not yet

learned calculus to solve for these similar questions, since my formulae do not require any

understanding of calculus or any other area of mathematics beyond high school algebra.

Considering that the formulae I find can decrease the number of operations one has to make and

thus shorten the calculation time for a given equation in the form of log𝑎 𝑓(𝑥) = 𝑔(𝑥), where

𝑓(𝑥) and 𝑔(𝑥) are polynomials, my solution can potentially be very useful for efficiently

solving these similar equations.

To proceed with my investigation, I will first make several assumptions that can simplify my

research and narrow my focus:

1. Despite the theory that two curves can intersect at infinity, I would not consider

intersection of log𝑎 𝑓(𝑥) and 𝑔(𝑥) at 𝑥 = ∞ or 𝑥 = −∞ to be an intersection.

2. All coefficients of 𝑓(𝑥) and 𝑔(𝑥) must be real numbers other than ± ∞.

3. log𝑎 𝑓(𝑥) has base 𝑎 > 1.

3

1. deg (f) = 0 or deg (g) = 0

I will discuss the two cases for which deg (f) = 0 or deg (g) = 0. When deg (f) = 0,

the two sides of the equation turn into a polynomial and a constant, which may be solved if

deg (g) < 5. When deg (g) = 0, the equation, under transformations, can still be written in

the form of a polynomial being equal to a constant term, which means that these two scenarios

are similar. These cases do not require a formula to determine whether log𝑎 𝑓(𝑥) and

𝑔(𝑥) intersect since they can both be simplified to a polynomial equation.

2. deg (f) = 1, deg (g) = 1

Let 𝑓(𝑥) = 𝑏𝑥 + c and 𝑔(𝑥) = 𝑑𝑥 + 𝑒, where 𝑏, c, 𝑑, 𝑒 ∈ R and 𝑏, 𝑑 ≠ 0.

Figure 1. 𝑙𝑜𝑔𝑎 𝑓(𝑥) = log (10x + 3)

4

The shape of this curve, as well as the shape of any other functions in the form of

log𝑎(𝑏𝑥 + 𝑐) for 𝑎 > 1 and 𝑏 > 0, is concave down. Each of these functions has one vertical

asymptote at the point where x = −𝑐

𝑏 and approaches ∞ when x approaches ∞. The slope of this

type of functions approximates ∞ as x approaches −𝑐

𝑏 and stays at 0 as x approaches ∞. For

𝑏 < 0, these functions remain concave down, but approaches ∞ when x approaches −𝑐

𝑏 .

For 𝑏 > 0 or 𝑏 < 0, 𝑔(𝑥) has to be strictly greater than log𝑎 𝑓(𝑥) for all real values of

x in the domain of log𝑎 𝑓(𝑥) in order to not intersect with log𝑎 𝑓(𝑥). Function 𝑔(𝑥) must

intersect with log𝑎 𝑓(𝑥) if 𝑔(𝑥) is set to be less than log𝑎 𝑓(𝑥) for all x in the domain of

log𝑎 𝑓(𝑥) since log𝑎 𝑓(𝑥) approaches −∞ as 𝑥 approaches −𝑐

𝑏.

When 𝑏 > 0, the leading coefficient of 𝑔(𝑥), d, must be greater than 0 such that no

intersection exists between log𝑎 𝑓(𝑥) and 𝑔(𝑥). This is due to the fact that log𝑎 𝑓(𝑥)

approaches ∞ when x approaches ∞ for 𝑏 > 0, and if 𝑑 < 0, then 𝑔(𝑥) approaches

−∞ when x approaches ∞. An intersection must exist in this case. Similarly, when 𝑏 < 0,

log𝑎 𝑓(𝑥) and 𝑔(𝑥) have no intersection only if 𝑑 < 0.

I will set restriction on the constant term of 𝑔(𝑥) to determine whether log𝑎 𝑓(𝑥) and

𝑔(𝑥) intersect. This way gives me the advantage of vertically translating 𝑔(𝑥) through

manipulating the value of e without changing the shape of 𝑔(𝑥), and has the nice property of

providing an inequality on the value of e.

Now, I suppose that the values for a, b, c, d, e are already given and it is my task to

determine whether an intersection exists between log𝑎 𝑓(𝑥) and 𝑔(𝑥).

5

In order to solve this question, I envision the function log𝑎 𝑓(𝑥) and 𝑔(𝑥) = 𝑑𝑥 + 𝑒 to be

tangent to each other, Since log𝑎 𝑓(𝑥) is concave down and 𝑔(𝑥) is linear, there is only one

intersection between these two functions, which is at the point of tangency. The reason that

log𝑎 𝑓(𝑥) and 𝑔(𝑥) do not intersect at any points other than the point of tangency is that, for x-

value greater or less than the x-value at the intersection point, the slope of 𝑔(𝑥) will be strictly

greater or less than the slope of log𝑎 𝑓(𝑥) respectively, assuming that b > 01. As a result, the

value of log𝑎 𝑓(𝑥) will be less than 𝑔(𝑥) for all values of x other than the point of tangency in

the domain of log𝑎 𝑓(𝑥) (“Rooftop Theorem”).

By increasing the value of e and thus translating 𝑔(𝑥) upwards, the intersection point

between these two functions disappears. Therefore, if e is greater than the value for which 𝑔(𝑥)

lies tangent to log𝑎 𝑓(𝑥), there is no intersection between these two functions. I will use this

property to set a restriction on e such that the two functions do not intersect.

From the formula for calculating the derivative of a logarithmic function,

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) =

𝑏

ln 𝑎 (𝑏𝑥 + 𝑐)

Suppose there exists a value 𝑥1 such that:

𝑑

𝑑𝑥log𝑎 𝑓(𝑥1) =

𝑑

𝑑𝑥𝑔(𝑥)

Then, we can differentiate both functions to determine 𝑥1:

1 The case in which 𝑏 < 0 is similar to 𝑏 > 0, and it can be shown that log𝑎 𝑓(𝑥) does not

intersect 𝑔(𝑥) other than the point of tangency.

6

𝑏

ln 𝑎 ∙ (𝑏𝑥1 + 𝑐)= 𝑑

𝑏

ln 𝑎 ∙ 𝑑= 𝑏𝑥1 + 𝑐

𝑥1 =1

ln 𝑎 ∙ 𝑑−

𝑐

𝑏

If log𝑎 𝑓(𝑥1) < 𝑔(𝑥1), there would be no intersection between the two functions:

log𝑎[ 𝑏(𝑥1) + 𝑐] < 𝑑(𝑥1) + 𝑒

log𝑎 [𝑏 (1

ln 𝑎 ∙ 𝑑−

𝑐

𝑏) + 𝑐] < 𝑑 (

1

ln 𝑎 ∙ 𝑑−

𝑐

𝑏) + 𝑒

𝑒 > log𝑎(𝑏

ln 𝑎 ∙ 𝑑) −

1

ln 𝑎+

𝑐𝑑

𝑏

In this case, for 𝑓(𝑥) and 𝑔(𝑥) coefficients a, b, c, d, e that satisfy this inequality, there is

no real solution for log𝑎 𝑓(𝑥) = 𝑔(𝑥).

3. deg (f) = 2, deg (g) = 1

Let 𝑓(𝑥) = 𝑏𝑥2 + 𝑐𝑥 + 𝑑 and 𝑔(𝑥) = 𝑒𝑥 + 𝑓, where b, c, d, e, f ∈ R and b, e ≠ 0.

There are two different behaviour of log𝑎 𝑓(𝑥) for 𝑏 > 0 and 𝑏 < 0 . Here are two

examples of log𝑎 𝑓(𝑥) for each of the two cases:

7

Figure 2. 𝑙𝑜𝑔𝑎 𝑓(𝑥) = 𝑙𝑜𝑔( 𝑥2 + 10𝑥 + 1)

Figure 3. 𝑙𝑜𝑔𝑎 𝑓(𝑥) = 𝑙𝑜𝑔(− 𝑥2 + 10𝑥 + 1)

For 𝑏 > 0, log𝑎 𝑓(𝑥) approaches ∞ when x approaches ∞. For 𝑏 < 0, log𝑎 𝑓(𝑥) has

two vertical asymptotes at 𝑥 = −𝑐±√𝑐2−4𝑏𝑑

2𝑏, because 𝑓(𝑥) is positive between its two roots.

It is helpful to note that 𝑥 ∈ ∅ if 𝑐2 − 4𝑏𝑑 < 0, and consequently there would be no intersection

8

between log𝑎 𝑓(𝑥) and 𝑔(𝑥). In these two scenarios, the shapes of log𝑎 𝑓(𝑥) greatly differ

from each other in their end behaviour.

When 𝑏 > 0, there must be an intersection between log𝑎 𝑓(𝑥) and 𝑔(𝑥), since 𝑔(𝑥)

approaches ∞ on one infinity side and −∞ on the other, whereas log𝑎 𝑓(𝑥) approaches ∞ on

both sides. Thus, there may exist no intersection between these two functions if and only if

𝑏 < 0.

Lemma 1. The shape of 𝑙𝑜𝑔𝑎 𝑓(𝑥) when b < 0 and 𝑐2 − 4𝑏𝑑 > 0 is concave down.

Due to the fact that the concavity of a function does not change as it undergoes horizontal or

vertical shift, translation, expansion, or compression, if log𝑎(𝑘 + 𝑏𝑥2) is concave down for a

given real number 𝑘 > 0, then log𝑎 𝑓(𝑥) has to be concave down. The function

log𝑎(𝑘 + 𝑏𝑥2) can be further transformed into log𝑎 𝑘 + log𝑎(1 +𝑏

𝑘𝑥2), and through horizontal

expansion or, this function transforms into log𝑎 𝑘 + log𝑎(1 − 𝑥2), for log𝑎 𝑘 as a real constant.

Given that a function is concave down if and only if its second derivative is less than 0, it is

important to calculate the second derivative of log𝑎(1 − 𝑥2). This is the end of the proof.

The tangent lines of concave down functions do not intersect with the functions at any points

other than their point of tangency. An upward vertical translation of 𝑔(𝑥), as a result, removes

the intersection between these two functions if 𝑔(𝑥) is tangent to log𝑎 𝑓(𝑥).

According to the formula for calculating the derivative of log𝑎 𝑓(𝑥),

9

𝑑

𝑑𝑥log𝑎 𝑏𝑥2 + 𝑐𝑥 + 𝑑 =

2𝑏𝑥 + 𝑐

ln 𝑎 ∙ (𝑏𝑥2 + 𝑐𝑥 + 𝑑)

Suppose there exists a number 𝑥1 in the domain of 𝑓(𝑥) such that:

𝑑

𝑑𝑥log𝑎 𝑓(𝑥1) =

𝑑

𝑑𝑥𝑔(𝑥1)

2𝑏𝑥1 + 𝑐

ln 𝑎 ∙ (𝑏𝑥12 + 𝑐𝑥1 + 𝑑)

= 𝑒

(2𝑏

ln 𝑎) 𝑥1 +

𝑐

ln 𝑎= 𝑏𝑒𝑥1

2 + 𝑐𝑒𝑥1 + 𝑑𝑒

𝑏𝑒𝑥12 + (𝑐𝑒 −

2𝑏

ln 𝑎) 𝑥1 + (𝑑𝑒 −

𝑐

ln 𝑎) = 0

𝑥12 + (

𝑐

𝑏−

2

ln 𝑎 ∙ 𝑒) 𝑥1 + (

𝑑

𝑏−

𝑐

𝑏𝑒 ln 𝑎) = 0

𝑥1 =(

2𝑒 ∙ ln 𝑎

−𝑐𝑏

) ± √(𝑐𝑏

−2

ln 𝑎 ∙ 𝑒)

2

− 4 (𝑑𝑏

−𝑐

𝑏𝑒 ln 𝑎)

2

𝑥1 =2𝑏 − 𝑐𝑒 ln 𝑎 ± √𝑐2𝑒2ln2𝑎 + 4𝑏2 − 4𝑏𝑑𝑒2ln2𝑎

2𝑏𝑒 ln 𝑎

There are two possible values of 𝑥1, but one of the two is not in the domain of log𝑎 𝑓(𝑥)

since the two functions would only intersect once in the given conditions. Let 𝑥1 that is within the

domain of log𝑎 𝑓(𝑥) be α.

In order for log𝑎 𝑓(𝑥) not to intersect with 𝑔(𝑥), log𝑎 𝑓(α) must be strictly less than 𝑔(α)

due to the fact that log𝑎 𝑓(𝑥) approaches −∞ at its vertical asymptotes. It follows that:

log𝑎( 𝑏𝛼2 + 𝑐𝛼 + 𝑑) < 𝑒𝛼 + 𝑓

𝑓 > log𝑎(𝑏𝛼2 + 𝑐𝛼 + 𝑑) − 𝑒𝛼

Therefore, if 𝑏 < 0 and 𝑐2 − 4𝑏𝑑 > 0 and the above inequality is satisfied, there exists no

solution for log𝑎 𝑓(𝑥) = 𝑔(𝑥) for deg (f) = 2 and deg (g) = 1.

10

4. deg (f) = 1, deg (g) = 2

Let 𝑓(𝑥) = 𝑏𝑥 + 𝑐 and 𝑔(𝑥) = 𝑑𝑥2 + 𝑒𝑥 + 𝑓, where b, c, d, e, f ∈ R and b, d ≠ 0.

Lemma 2. Considering the end behaviour of 𝑙𝑜𝑔𝑎 𝑓(𝑥) and 𝑔(𝑥), the leading coefficient

of 𝑔(𝑥), 𝑑, must be greater than 0 if there exists no intersection between the two functions.

If log𝑎 𝑓(𝑥) does not intersect 𝑔(𝑥), then log𝑎 𝑓(𝑥) must be either greater or smaller than

𝑔(𝑥) for all 𝑥 in the domain of log𝑎 𝑓(𝑥). However, since, if 𝑑 < 0, 𝑔(𝑥) would reach −∞

as x approaches ∞, log𝑎 𝑓(𝑥) cannot be entirely below 𝑔(𝑥) for 𝑏 < 0. It cannot be entirely

above 𝑓(𝑥) either due to the fact that it approaches −∞ for some real value of 𝑥. This is the

end of the proof.

It is now clear to see that log𝑎 𝑓(𝑥) does not intersect 𝑔(𝑥) if and only if 𝑏 > 0. Ideally,

an upward vertical translation by a certain value on 𝑔(𝑥) can reduce the original two

intersections between the two functions to one point of tangency, such that any other upward

vertical translation of 𝑔(𝑥) would remove all intersections between them. My goal is to find

such a value.

Lemma 3. The intersection between the concave up function 𝑔(𝑥) and the concave down

function 𝑙𝑜𝑔𝑎 𝑓(𝑥) is only one point if and only if the two functions are tangent to each other.

11

Let 𝑥1 be the point of tangency between the two functions. Then, 𝑑

𝑑𝑥log𝑎 𝑓(𝑥1) =

𝑑

𝑑𝑥𝑔(𝑥1). For ∀ 𝑥: 𝑥 > 𝑥1,

𝑑

𝑑𝑥𝑔(𝑥) >

𝑑

𝑑𝑥𝑔(𝑥1), while

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) <

𝑑

𝑑𝑥log𝑎 𝑓(𝑥1) due to

the properties of concave up and concave down functions. Thus, 𝑑

𝑑𝑥𝑔(𝑥) >

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) for

∀ 𝑥: 𝑥 > 𝑥1. Similarly, we can deduce that 𝑑

𝑑𝑥𝑔(𝑥) <

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) for ∀ 𝑥 < 𝑥1. Since

log𝑎 𝑓(𝑥1) = 𝑔(𝑥1), 𝑔(𝑥) > log𝑎 𝑓(𝑥) for ∀𝑥: 𝑥 ≠ 𝑥1. This is the end of the proof.

In order to locate that potential point of tangency, it is important to first locate the point at

which 𝑑

𝑑𝑥log𝑎 𝑓(𝑥) =

𝑑

𝑑𝑥𝑔(𝑥). Suppose the two functions have the same slope at 𝑥1:

𝑑

𝑑𝑥log𝑎 𝑓(𝑥1) =

𝑑

𝑑𝑥𝑔(𝑥1)

𝑏

ln 𝑎 ∙ (𝑏𝑥1 + 𝑐)= 2𝑑𝑥1 + 𝑒

𝑏

ln 𝑎= (2𝑑𝑥1 + 𝑒)(𝑏𝑥1 + 𝑐)

2𝑏𝑑𝑥12 + (2𝑐𝑑 + 𝑏𝑒)𝑥1 + 𝑒𝑐 −

𝑏

ln 𝑎= 0

𝑥12 + (

𝑐

𝑏+

𝑒

2𝑑) 𝑥1 + (

𝑐𝑒

2𝑏𝑑−

1

2𝑑 ln 𝑎) = 0

𝑥1 =−

𝑐𝑏

−𝑒

2𝑑± √(

𝑐𝑏

+𝑒

2𝑑)2 + 4(

12𝑑 ln 𝑎

−𝑐𝑒

2𝑏𝑑)

2

𝑥1 =−2𝑏𝑐 ln 𝑎 − 𝑏𝑒 ln 𝑎 ± √4𝑏2𝑐2ln2𝑎 + 𝑏2𝑒2ln2𝑎 + 8𝑏2𝑑 ln 𝑎 − 4𝑏𝑐𝑑𝑒ln2𝑎

4𝑏𝑑 ln 𝑎

There are two possible values of 𝑥1 , but one is not in the domain of log𝑎 𝑓(𝑥) since

log𝑎 𝑓(𝑥) and 𝑔(𝑥) intersect only once if 𝑑 < 0 and the two functions are tangent to each other.

Let the value 𝑥1 that is within the domain of log𝑎 𝑓(𝑥) to be α.

12

log𝑎 𝑓(𝑥) must be strictly less than 𝑔(𝑥), which can be satisfied by letting 𝑔(𝑥) be greater

than log𝑎 𝑓(𝑥) at 𝑥1. An inequality between the two functions follows:

log𝑎(𝑏𝛼 + 𝑐) < 𝑑𝛼2 + 𝑒𝛼 + 𝑓

𝑓 > log𝑎(𝑏𝛼 + 𝑐) − 𝑑𝛼2 − 𝑒𝛼

5. deg (f) = 2, deg (g) = 2

Let 𝑓(𝑥) = 𝑏𝑥2 + 𝑐𝑥 + 𝑑 and 𝑔(𝑥) = 𝑒𝑥2 + 𝑓𝑥 + 𝑔, where b, c, d, e, f, g ∈ R and

b, e ≠ 0.

5.1 𝒃 > 𝟎

When 𝑏 > 0, there are two possible shapes of log𝑎 𝑓(𝑥). One shape consists of two concave

down functions separated by a region in which 𝑓(𝑥) ≤ 0, as shown in Figure 2. It approaches

−∞ when x approaches −𝑐±√𝑐2−4𝑏𝑑

2𝑏. This shape is distinguished from the other shape by the

fact that it satisfies the inequality 𝑐2 ≥ 4𝑏𝑑. Let us denote the shape that satisfies the inequality

𝑐2 ≥ 4𝑏𝑑 to be the first shape, and the other shape to be the second shape.

5.1.1 The first shape

Lemma 4. log𝑎 𝑓(𝑥) consists of two concave down curves if 𝑏 > 0 and 𝑐2 ≥ 4𝑏𝑑.

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It is useful to note that the concavity of log𝑎 𝑓(𝑥) is preserved after horizontal and vertical

transformations. As a result, we can rewrite log𝑎 𝑓(𝑥) as log(𝑥2 + 𝑚𝑥 + 𝑛), with 𝑚2 ≥ 4𝑛,

and then further transform the function to log[(𝑥 +1

2𝑚)

2

+𝑛−1

4𝑚2]. The resulting function, after

horizontal shift, is log(𝑥2 − 𝑘) for some positive real number k, which is equivalent to

log (1

𝑘𝑥2 − 1) + log 𝑘. After vertical translation and horizontal expansion or compression, this

function ultimately becomes log(𝑥2 − 1). 𝑑2

𝑑𝑥2 log(𝑥2 − 1) = 𝑟(−2𝑥2−2)

(𝑥2+1)2 for some positive real

number r. From this equation, we can see that 𝑑2

𝑑𝑥2 log(𝑥2 − 1) < 0. Therefore, log(𝑥2 − 1)

consists of two concave down curves, which implies that the first shape of log𝑎 𝑓(𝑥) also has

two concave down curves. This is the end of the proof.

The first shape that consists of two separate concave down curves cannot be discussed using

the aforementioned vertical translation of 𝑔(𝑥). Since this shape has two separate curves, 𝑔(𝑥)

may be tangent to one curve of this shape but intersect at two points with the other curve, such

that when vertical translation is applied to 𝑔(𝑥), there still exist intersections between log𝑎 𝑓(𝑥)

and 𝑔(𝑥). A slight modification is needed for this method.

First, I will determine the lines of reflection for the two curves. The function 𝑓(𝑥) reaches

minimum value at the vertex, where 𝑥 =−𝑐

2𝑏, and it is symmetrical on either side of the line of

reflection. This means that log𝑎 𝑓(𝑥) will also be symmetrical at 𝑥 =−𝑐

2𝑏.

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Lemma 5. a symmetrical pair of non-intersecting concave down curves and a symmetrical

concave up curve tangent to one of the concave down curves, both with reflection symmetry

parallel to the y-axis, intersect only at the tangent point if and only if the reflection line of the

concave up curve is on the same side as the tangent point according to the y-axis.

Without the loss of generality, we assume that the reflection line of the symmetrical concave

up curve is on the left side of the reflection line of the two concave down curves. By applying a

horizontal shift of the right side concave down curve to the left by a certain number of units, we

can create another concave down curve such that it is symmetrical to the left side concave down

curve at the reflection line of the concave up curve. If the concave up curve is tangent to the left

side concave down curve, then it must also be tangent to the new right side concave down curve

after the horizontal shift. Due to the fact that the concave up curve would intersect the right side

concave down curve from the left, it would intersect with the transformed concave down curve

for two times before it intersects with the original right side concave down curve. However, since

the concave up curve intersects with the transformed concave down curve only once, there would

be no intersection between the concave up curve and the right side concave down curve. This is

the end of the proof.

This lemma indicates that if we choose the tangent point to be on the same side as the vertex

of 𝑔(𝑥) with respect to the reflection line of log𝑎 𝑓(𝑥), then there will only be one point of

intersection between the two functions. This fact proves to be extremely useful by providing

information about the lower bound on the value of the constant term g in 𝑔(𝑥) such that

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log𝑎 𝑓(𝑥) and 𝑔(𝑥) do not intersect.

5.1.2 The second shape

The second shape is a continuous function that resembles the first shape, but is different in

that it is not entirely concave down. Here is an example of the second shape:

Figure 4. 𝑙𝑜𝑔𝑎 𝑓(𝑥) = 𝑙𝑜𝑔(𝑥2 + 5𝑥 + 12)

In the second shape, there exist two inflection points at which the slope of the function

changes from decreasing to increasing and from increasing to decreasing. The situation in which

the second shape would occur is when the equation 𝑓(𝑥) = 0 has no real solution.

Essentially, there needs to be a minimum value on the constant value of 𝑔(𝑥) such that

log𝑎 𝑓(𝑥) does not intersect with 𝑔(𝑥).

The difficulty of this case lies in the fact that the function log𝑎 𝑓(𝑥) will no longer be just a

concave down curve and cannot be dealt with using the same method. To simplify this problem, I

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will apply case analysis by examining the two distinct shapes of log𝑎 𝑓(𝑥) for 𝑏 > 0

individually.

The second shape is a continuous, non-concave down function with two inflection points,

and we can prove this fact through algebraic manipulation.

Lemma 6. 𝑙𝑜𝑔𝑎 𝑓(𝑥) = 𝑏𝑥2 + 𝑐𝑥 + 𝑑 has two inflection points if b > 0 and 𝑐2 < 4𝑏𝑑,

with b, c, d being real numbers.

log𝑎 𝑓(𝑥) does not change its general shape after horizontal or vertical shift, translation,

expansion, or compression. The function can be reduced to log𝑎(1 + 𝑥2). 𝑑2

𝑑𝑥2 log𝑎(1 + 𝑥2) =

2−2𝑥2

ln 𝑎∙(1+𝑥2)2, which equals to 0 only if 2 − 2𝑥2 = 0. Since the numerator, (2 − 2𝑥2), is a

quadratic polynomial, there are at most two solutions to 𝑑2

𝑑𝑥2 log𝑎(1 + 𝑥2) = 0. Thus, log𝑎 𝑓(𝑥)

has at most 2 inflections points.

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) = 0 at the reflection line2, and gradually approaches 0 as x approaches ∞.

Therefore, there must exist at least one point on the right side of the reflection line such that the

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) changes from increasing to decreasing. Due to the fact that the second shape of

log𝑎 𝑓(𝑥) is symmetrical, there exist at least two inflection points for log𝑎 𝑓(𝑥). Therefore, the

second shape of log𝑎 𝑓(𝑥) has two inflection points, one on each side of the reflection line. This

is the end of the proof.

2 The slope of a symmetrical function that is continuous on (−∞, ∞) and differentiable on (−∞, ∞) equals 0 at the reflection line,

17

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) approaches 0 as x approaches ± ∞, and has a rotational symmetry around the

point on log𝑎 𝑓(𝑥) at x =−𝑐

2𝑏, where the slope of log𝑎 𝑓(𝑥) = 0. An example of this derivative

is shown on the figure below:

Figure 5. 𝑦 =20𝑥+5

10𝑥2+5𝑥+1

Note that the derivative of 𝑔(𝑥) is a line with a positive slope.

As mentioned previously, the tangent point that will be taken must be on the same side as the

vertex of 𝑔(𝑥) according to the reflection point at 𝑥 =−𝑐

2𝑏 . This means that, if the tangent point

taken is at the left side of 𝑥 =−𝑐

2𝑏, then the derivative of 𝑔(𝑥) must intersect with the x-axis left

to 𝑥 =−𝑐

2𝑏 as well.

Since the derivative of log𝑎 𝑓(𝑥) has rotational symmetry around the point on log𝑎 𝑓(𝑥) at

𝑥 =−𝑐

2𝑏, and the set of all lines with positive slope is unchanged after an 180° rotation around that

18

point, without the loss of generality, we can assume that the vertex of 𝑔(𝑥) is on the left side of

𝑥 =−𝑐

2𝑏.

From the fact that the derivative of 𝑔(𝑥) is a line with a positive slope, we can deduce that

the derivative of 𝑔(𝑥) will only intersect once with the derivative of log𝑎 𝑓(𝑥) on the left side

of 𝑥 =−𝑐

2𝑏. This finding is very useful, since if there are two intersections between the derivative

of 𝑔(𝑥) and log𝑎 𝑓(𝑥) at the left hand side of 𝑥 =−𝑐

2𝑏, then it is more difficult to determine

which tangent point to use as the boundary for the constant term, g, in 𝑔(𝑥).

It is important to note that, if log𝑎 𝑓(𝑥) and 𝑔(𝑥) are tangent to each other and the vertex

of 𝑔(𝑥) and the tangent point are on the same side with respect to 𝑥 =−𝑐

2𝑏, then there will be no

intersection between log𝑎 𝑓(𝑥) and 𝑔(𝑥) on the other side of 𝑥 =−𝑐

2𝑏.

What we have determined here is that the case of the second shape is congruent with the case

of the first one, and that it can be solved using the same method.

Due to the occurrence of the second shape of log𝑎 𝑓(𝑥) for 𝑏 > 0, the function 𝑔(𝑥) is

allowed to open downwards such that there exists no intersection between log𝑎 𝑓(𝑥) and 𝑔(𝑥).

In other words, e, the leading coefficient of 𝑔(𝑥), can be negative. The reason is that the second

shape of log𝑎 𝑓(𝑥) does not approach negative infinity for any real values of x, and thus allows

𝑔(𝑥) to approach −∞ when x approaches ± ∞.

Using the same approach as before, we first try to determine the values of x for which the

derivative of log𝑎 𝑓(𝑥) and the derivative of 𝑔(𝑥) are the same. This question simplifies to

solving a polynomial equation with degree 3, which has a maximum of three real solutions. In

19

other words, the derivative of log𝑎 𝑓(𝑥) and the derivative of 𝑔(𝑥) intersects at most three

times.

There are three possibilities for the intersections between the derivatives of the two

functions: one intersection, two intersections including a tangent point, and three intersections.

If there exists only one solution to the polynomial equation, which means that there exists

one intersection between the derivatives, we should choose that point as the tangent point

between log𝑎 𝑓(𝑥) and 𝑔(𝑥) and determine a restriction on the value of g, the constant term of

𝑔(𝑥) such that there exists no intersection between the two functions. The reason why this

method works is that to the left of the intersection of the two derivatives, the derivative of 𝑔(𝑥)

is greater than that of log𝑎 𝑓(𝑥), and to the right of the intersection, that of 𝑔(𝑥) is less than

that of log𝑎 𝑓(𝑥), which guarantees that 𝑔(𝑥) will be less than log𝑎 𝑓(𝑥) for all values of x

other than the value of x at the tangent point.

It is possible for the two derivatives to have two intersections, but one of the intersections

must be a tangent point between the derivatives. If log𝑎 𝑓(𝑥) = 𝑔(𝑥) at this tangent point

between the derivatives, the two functions cross each other at this point even though their slopes

are equal. This point is also called the inflection point of the two tangent curves. This point can

be disregarded in our investigation, since it does not provide us with information about the

restriction on the constant term g for log𝑎 𝑓(𝑥) and 𝑔(𝑥) to not intersect. The way to see if an

intersection between the derivatives is a tangent point is by calculating the second derivatives of

log𝑎 𝑓(𝑥) and 𝑔(𝑥). If the second derivatives are equal to each other at the intersection, then

20

this intersection is indeed a tangent point. After knowing which one is the inflection point, we

would just use the other intersection and employ the same method previously mentioned.

If there are three intersections between the two derivatives, then we call the x-coordinates of

the intersections as 𝑥1, 𝑥2, and 𝑥3, with 𝑥1 > 𝑥2 > 𝑥3. Since there is no tangent point

between the derivatives in these three intersections, 𝑑

𝑑𝑥log𝑎 𝑓(𝑥) >

𝑑

𝑑𝑥𝑔(𝑥) for 𝑥 between 𝑥2

and 𝑥3, and 𝑑

𝑑𝑥log𝑎 𝑓(𝑥) <

𝑑

𝑑𝑥𝑔(𝑥) for 𝑥 between 𝑥1 and 𝑥2.

To find the tangent point between log𝑎 𝑓(𝑥) and 𝑔(𝑥) such that it is the only point of

intersection, we first assume each of the three points, 𝑥1, 𝑥2, and 𝑥3 to be that tangent point.

Through calculations same to the ones mentioned above, we would obtain three maximum values

on the value of g such that log𝑎 𝑓(𝑥) and 𝑔(𝑥) do not intersect. At this point, we choose the

least of the three values as the restriction on g.

5.2 𝒃 < 𝟎

The shape of log𝑎 𝑓(𝑥) when 𝑏 < 0 and 𝑐2 − 4𝑏𝑑 > 0 is concave down3. 𝑔(𝑥) can be

either a concave up or concave down quadratic function.

If 𝑔(𝑥) is concave up, which implies that 𝑒 > 0, then there would be only one intersection

between log𝑎 𝑓(𝑥) and 𝑔(𝑥) if the two functions are tangent to each other4. It follows that:

𝑑

𝑑𝑥log𝑎 𝑓(𝑥) =

𝑑

𝑑𝑥𝑔(𝑥)

2𝑏𝑥 + 𝑐

ln 𝑎 ∙ (𝑏𝑥2 + 𝑐𝑥 + 𝑑)= 2𝑒𝑥 + 𝑓

3 Refer to the proof of the lemma 1 on page 8 4 Refer to the proof of the lemma 5 on page 10

21

(2𝑏

ln 𝑎) 𝑥 + (

𝑐

ln 𝑎) = 2𝑏𝑒𝑥3 + (2𝑐𝑒 + 𝑏𝑓)𝑥2 + (2𝑑𝑒 + 𝑐𝑓)𝑥 + 𝑑𝑓

2𝑏𝑒𝑥3 + (2𝑐𝑒 + 𝑏𝑓)𝑥2 + (2𝑑𝑒 + 𝑐𝑓 −2𝑏

ln 𝑎) 𝑥 + (𝑑𝑓 −

𝑐

ln 𝑎) = 0

By applying the cubic formula, we can find at least one and at most three solutions. Only one

value would be within the domain of log𝑎 𝑓(𝑥) due to the fact that there is at most one point for

which a concave up and a concave down function has the same slope. Let that value be α.

log𝑎(𝑏𝛼2 + 𝑐𝛼 + 𝑑) < 𝑒𝛼2 + 𝑓𝛼 + 𝑔

𝑔 > log𝑎(𝑏𝛼2 + 𝑐𝛼 + 𝑑) − 𝑒𝛼2 − 𝑓𝛼

There would be no intersection between the two functions if this inequality is satisfied with

the given conditions.

6. Further observation

Lemma 7. Two functions, 𝑓(𝑥) and 𝑔(𝑥), that are differentiable on the interval of

(−∞, ∞) and continuous on the interval of (−∞, ∞) do not intersect only if

∃ 𝑥1 ∈ R: 𝑑

𝑑𝑥𝑓(𝑥1) =

𝑑

𝑑𝑥𝑔(𝑥1)

Given that 𝑓(𝑥) and 𝑔(𝑥) are two non-intersecting continuous and differentiable functions

with x ∈ R, without the loss of generality, we can assume that 𝑓(𝑥) is greater than 𝑔(𝑥) for all

real values of x. By applying downward vertical translation on 𝑓(𝑥), we can guarantee that 𝑓(𝑥)

and 𝑔(𝑥) would intersect. Suppose that after translating 𝑓(𝑥) n units down, 𝑓(𝑥) and 𝑔(𝑥)

intersect for the first time. We can show that the intersections must be tangent points, by

22

contradiction. If one intersection is not a tangent point, then either on the left side or the right side

of that intersection there must be an interval of x-values such that 𝑓(𝑥) < 𝑔(𝑥). This means that

if we can translate 𝑓(𝑥) up by a certain unit such that 𝑓(𝑥) still intersects with 𝑔(𝑥),

contradicting the assumption that 𝑓(𝑥) first intersects with 𝑔(𝑥) after its translation by n units

downward. This is the end of the proof.

This is an observation that I have made from my research, and would have great contribution

to future investigation. The implication is that if I determine that it is possible for two functions

to not intersect, I only have to locate the x-values for which the two functions have the same

slope, and determine for which x-value would place the greatest restriction on the constant term

for one of the function, significantly simplifying my future research.

7. Conclusion

The purpose of my research is to find a more efficient way of determining the existence of

intersections between a logarithmic function and a polynomial. I have discussed four cases and

provided solutions to each of the cases with respect to my research question.

The method I have employed, which is analyzing the behaviour of functions and finding

tangent points between the two functions, unfortunately becomes ineffective when the degrees of

𝑓(𝑥) and 𝑔(𝑥) begin to increase. The reason comes from the fact stated in the Abel-Ruffini

23

Theorem: there exists no general algebraic solution to a polynomial equation with degree greater

than or equal to 5 (Żolądek, 2000). It is important to note that this theorem does not claim that there

is no solution to polynomial equations of degree greater than or equal to 5.

The original way for locating a point at which the slopes of log𝑎 𝑓(𝑥) and 𝑔(𝑥) are equal

to each other would not work with this restriction. To let 𝑑

𝑑𝑥𝑓(𝑥) be equal to

𝑑

𝑑𝑥𝑔(𝑥) would then

turn into a polynomial equation with degree of deg (f) + deg (g) − 1, and it must be strictly

less than 5. Equivalently, deg (f) + deg (g) must be smaller than 6 in order for our original

method to work. I left out 6 cases that may be solved using this method. I would not discuss them,

since they can all be solved using the similar way of analyzing functions and finding tangent points

between them.

There are other ways of solving a polynomial equation. One of which is the Newton’s

method, which gives good approximation to the zeroes of a function. Another way is by graphing

the polynomial and see if it intersects with the x-axis. However, both of them can be time-

consuming.

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Bibliography

Rooftop Theorem for Concave down Functions. University of Southern California, Santa

Barbara. Retrieved from

http://www.econ.ucsb.edu/~tedb/Courses/GraduateTheoryUCSB/Rooftop.pdf

Żolądek, Henryk. (2000). The Topological Proof of Abel-Ruffini Theorem. Journal of the

Juliusz Schauder Center, 16, 253-265.