NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Difference Quotients
Lesson

A difference quotient is a mathematical expression that is commonly used to establish an average rate of change between two points belonging to a smoothly varying function.

Suppose $a$a and $b$b are two values in the domain of a function $f$f. The function values at $a$a and $b$b are $f(a)$f(a) and $f(b)$f(b). The difference between $a$a and $b$b, given by $b-a$ba, corresponds to the difference between the function values, given by $f(b)-f(a)$f(b)f(a)

The difference quotient $\frac{f(b)-f(a)}{b-a}$f(b)f(a)ba expresses the relative change of the function value for the given change in the domain variable.

When this is pictured in the form of a graph, we may speak of this as the rise over the run, meaning the change in the function, given by the vertical difference between the points, for a given change in the domain variable, given by the horizontal difference between the points.

Equivalently, we can say that the difference quotient expresses the gradient of the line joining the points $\left(a,f(a)\right)$(a,f(a)) and $\left(b,f(b)\right)$(b,f(b)).

The difference quotient expresses the average rate of change of a function in moving from, say, $x=a$x=a to $x=b$x=b. When $b$b is close to $a$a, the difference quotient is near in value to the instantaneous rate of change of the function at $a$a. On a graph, we see that the secant line joining $\left(a,f(a)\right)$(a,f(a)) and $\left(b,f(b)\right)$(b,f(b)) becomes indistinguishable from the tangent line at $a$a when $b$b moves very close to $a$a.

We say that the gradient of the tangent at $a$a is the limit as $b\rightarrow a$ba of the difference quotient $\frac{f(b)-f(a)}{b-a}$f(b)f(a)ba. The gradient of the tangent is the instantaneous rate of change of the function at $a$a.

To make more general statements about difference quotients and gradients, we consider a point $x$x in the domain of a function and another point $x+h$x+h that is close to $x$x when $h$h is small. The difference quotient now takes the form $\frac{f(x+h)-f(x)}{x+h-x}$f(x+h)f(x)x+hx, which is just

$\frac{f(x+h)-f(x)}{h}$f(x+h)f(x)h

Observe that if we let $h=0$h=0, the quotient becomes $\frac{0}{0}$00, which is meaningless. And yet, if we let $h$h approach $0$0 ever more closely, we often see that the difference quotient approaches a definite value. This discovery is the basis for differential calculus.

#### Example 1

Let $y$y be a function of $x$x where $x$x can be any real number. The function is given by the formula $y(x)=\frac{x}{2}+3$y(x)=x2+3. Find an expression for the difference quotient at any point $x$x in the domain.

According to the function definition, we have $y(x+h)=\frac{x+h}{2}+3$y(x+h)=x+h2+3. So, we can form the difference quotient

 $\frac{y(x+h)-y(x)}{h}$y(x+h)−y(x)h​ $=$= $\frac{\frac{x+h}{2}+3-(\frac{x}{2}+3)}{h}$x+h2​+3−(x2​+3)h​ $=$= $\frac{\frac{x}{2}+\frac{h}{2}+3-\frac{x}{2}-3}{h}$x2​+h2​+3−x2​−3h​ $=$= $\frac{\frac{h}{2}}{h}$h2​h​ $=$= $\frac{1}{2}$12​

We see that the gradient of a line joining any two points belonging to this function is always $\frac{1}{2}$12, as we would expect for a function whose graph is a straight line.

#### Example 2

Find the general difference quotient expression for the function $f(x)=x^2-x+2$f(x)=x2x+2. Use it to find the gradients of secant lines at $x=1$x=1 when $h=1,0.1,0.01,0.001$h=1,0.1,0.01,0.001.

We have, $f(x+h)=(x+h)^2-(x+h)+2$f(x+h)=(x+h)2(x+h)+2. Therefore

 $\frac{f(x+h)-f(x)}{h}$f(x+h)−f(x)h​ $=$= $\frac{(x+h)^2-(x+h)+2-(x^2-x+2)}{h}$(x+h)2−(x+h)+2−(x2−x+2)h​ $=$= $\frac{x^2+2xh+h^2-x-h+2-x^2+x-2}{h}$x2+2xh+h2−x−h+2−x2+x−2h​ $=$= $\frac{2xh+h^2-h}{h}$2xh+h2−hh​ $=$= $\frac{h(2x+h-1)}{h}$h(2x+h−1)h​ $=$= $2x+h-1$2x+h−1

The secant lines at $x=1$x=1, therefore, have gradients $1+h$1+h. Thus, when $h=1$h=1 the secant line has gradient $2$2; when $h=0.1$h=0.1, the gradient is $1.1$1.1; when $h=0.01$h=0.01, the gradient is $1.01$1.01 and when $h=0.001$h=0.001, the secant line has gradient $001$001.

#### Worked Examples

##### Question 1

The population $f\left(x\right)$f(x) of bacteria present in some food $x$x minutes after it was left out of the fridge is given by $f\left(x\right)=x^2+3$f(x)=x2+3.

The graph of the function has been provided.

1. What is the population of bacteria $1$1 minute after leaving the food out of the fridge?

2. What is the population of bacteria $2$2 minutes after leaving the food out of the fridge?

3. What is the average rate of increase of bacteria between $1$1 and $2$2 minutes?

4. We now want to generalise the average rate of change between any two times.

If the number of bacteria at $x=a$x=a minutes is $f\left(a\right)$f(a) and the number of bacteria at $x=a+h$x=a+h minutes is $f\left(a+h\right)$f(a+h), fill in the gaps to form an expression for the average rate of change over this interval of time.

 Total change in the quantity $=$= $\editable{}-f\left(a\right)$−f(a) Total change in time $=$= $\editable{}-a$−a Average rate of change $=$= $\frac{\editable{}-\editable{}}{\editable{}}$−​

##### question 2

Consider the function $f\left(x\right)=6x+7$f(x)=6x+7.

1. Calculate the difference quotient for $f\left(x\right)$f(x).

2. Using your answer to the last part, write down the value of the difference quotient when $x=6$x=6 and $x+h=7$x+h=7.

##### question 3

Consider the function $f\left(x\right)=3x^2+5x+4$f(x)=3x2+5x+4. The points $A$A and $B$B have coordinates $\left(x,f\left(x\right)\right)$(x,f(x)) and $\left(x+h,f\left(x+h\right)\right)$(x+h,f(x+h)) respectively.

1. Find $f\left(x+h\right)$f(x+h). Expand the expression fully.

2. Find $f\left(x+h\right)-f\left(x\right)$f(x+h)f(x). Fully factorise your final answer.

3. Find $\frac{f\left(x+h\right)-f\left(x\right)}{h}$f(x+h)f(x)h. Expand your final answer fully.

4. The points $A$A and $B$B have coordinates $\left(x,f\left(x\right)\right)$(x,f(x)) and $\left(x+h,f\left(x+h\right)\right)$(x+h,f(x+h)) respectively.

The difference quotient of $f\left(x\right)$f(x) represents the gradient of which one of the following lines and segments?

A

B

C

D

A

B

C

D

### Outcomes

#### M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

#### 91578

Apply differentiation methods in solving problems