Comparing Functions (quad, cubic, exp, linear)

We sometimes need to compare the behaviour of two different functions, perhaps over a particular domain interval.

We may do this by looking at the graphs of the functions, by comparing tables of values of the functions or by examining the algebraic representations of the functions.

Questions we may wish to answer in comparing two functions include such things as

• whether one function is always greater than or less than the other,
• whether the graphs intersect and if so how often,
• whether one or both functions are strictly increasing or decreasing on an interval,
• how the function values compare at selected points in the domain,
• how the gradients differ, and
• how the maximum or minimum values compare.

Example 1

Of the two functions $f(x)=\frac{1}{2}e^x-1$f(x)=12​ex−1 and $g(x)=x-\frac{1}{2}$g(x)=x−12​, which has the greater value at $x=0$x=0, and which has the greater value at $x=1$x=1?

The question is easy to resolve by looking at the graphical representations.

 

The algebraic function definitions tell us that $f(0)=\frac{1}{2}e^0-1=-\frac{1}{2}$f(0)=12​e0−1=−12​ and that $g(0)=0-\frac{1}{2}=-\frac{1}{2}$g(0)=0−12​=−12​. So, the functions are equal at $x=0$x=0.

At $x=1$x=1, we have $f(1)=\frac{1}{2}e^1-1\approx0.359$f(1)=12​e1−1≈0.359. But $g(1)=1-\frac{1}{2}=\frac{1}{2}$g(1)=1−12​=12​. So, $g(1)>f(1)$g(1)>f(1).

 

Example 2

We know that $g(1)>f(1)$g(1)>f(1) for $f$f and $g$g defined as in the previous example. Compare $g(2)$g(2) and $f(2)$f(2) and consider what conclusion can be reached.

We have $f(2)=\frac{1}{2}e^2-1>\frac{1}{2}.\left(\frac{5}{2}\right)^2-1=\frac{17}{8}$f(2)=12​e2−1>12​.(52​)2−1=178​ and $g(2)=2-\frac{1}{2}=\frac{3}{2}$g(2)=2−12​=32​. So, $f(2)>g(2)$f(2)>g(2).

The two functions, $f$f and $g$g, are continuous. So, somewhere between $x=1$x=1 and $x=2$x=2 their graphs must cross and therefore there must exist an $x$x such that $f(x)=g(x)$f(x)=g(x). 

We know that this $x$x exists but it is not easy to find what it is.

 

Example 3

Considering again the two functions $f$f and $g$g as defined in Example 1, is there a value of $x$x such that the two graphs have the same slope? That is, can we find an $x=x_0$x=x0​ such that the functions are changing at the same rate in the vicinity of $x_0$x0​?

At $x=0$x=0, the two graphs intersect and the gradient of the function $f$f is less than the gradient of $g$g. However, at the value of $x$x where the graphs intersect again, the gradient of $f$f is greater than the gradient of $g$g. Since the gradients vary smoothly between these two points, it must be that there is an $x$x in between where the gradients are the same.

This is a theorem that is encountered and proved in calculus.

 

Example 4

Consider the functions $y(x)=x^3+x$y(x)=x3+x and $u(x)=x^2+1$u(x)=x2+1 on the interval $[-1,1]$[−1,1]. Compare the values of both functions at the end-points of the interval and also at the point $x=0$x=0.

We could display the results of this investigation in a table.

$x$x	$y(x)$y(x)	$u(x)$u(x)
$-1$−1	$-2$−2	$2$2
$0$0	$0$0	$1$1
$1$1	$2$2	$2$2

After assembling this table, further questions arise: Does $u(x)$u(x) ever go below $1$1? Is $y(x)\le u(x)$y(x)≤u(x) for every $x$x in the interval $[-1,1]$[−1,1]?

From the function definition, $u(x)=x^2+1$u(x)=x2+1 we can see that the $x^2$x2 term can never be less than $0$0. So, $u(x)$u(x) can never be less than $1$1.

To answer the second question, we could consider the functions over the negative and positive parts of the domain separately. It is clearly true that $y(x)y(x)<u(x) when $x$x is negative because $x^3+x$x3+x is negative while $x^2+1$x2+1 is positive. This would leave just the interval $[0,1]$[0,1] to think about.

Alternatively, thinking about the whole interval $[-1,1]$[−1,1], we write the inequality $y(x)\le u(x)$y(x)≤u(x) and try to simplify it to find out for which values of $x$x it is true. We have

$x^3+x$x3+x	$<$<	$x^2+1$x2+1
$\therefore$∴ $x\left(x^2+1\right)$x(x2+1)	$<$<	$x^2+1$x2+1
$\therefore x$∴x	$<$<	$1$1

We were able to divide through by $x^2+1$x2+1 in the second line without changing the direction of the inequality because this expression is always positive.

We conclude that $y(x)\le u(x)$y(x)≤u(x) for all values of $x$x less than $1$1.

 

Examples

QUESTION 1

QUESTION 2

QUESTION 3