# Difference quotient: applications of functions

## What is the Difference Quotient?

Before we define the deviation quotient and the deviation quotient formula, it is crucial to first understand the definition of derivatives .
In the most basic sense, a derivative is a measure of a function ‘s rate of change. That is, the instantaneous pace of change at a given compass point. In regulate to measure this value, we use what is called a tangent line, and meter its gradient to given the best potential linear approximation at a single detail .
In this article, however, we wo n’t be looking at derivatives in particular. rather, we will be looking at the deviation quotient, which is a stepping stone to calculating derivatives of functions. The remainder quotient allows us to compute the gradient of secant lines. A secant line is closely the same as a tangent argumentation, but it rather goes through at least two points on a officiate .
In order to understand what the deviation quotient is, it helps to visualize two points on a line like the follow image :

Visualize two points on a line to understand difference quotient

If we recall the definition of gradient, we can use this definition to find the “ secant line ” going through two points. furthermore, if we use the variable “ planck’s constant ” to represent the difference between the two x-values, we start to see a recipe develop :

Develop a formula using definition of slope, secant line and variable h

last, with some delete of terms, we can arrive at the very definition of the difference quotient .

Cancel terms to get the definition of difference quotient

therefore, the formula for the difference quotient is :

f(a+h)−f(a)h\frac{f(a+h)-f(a)}{h}

h

f

(

a

+

h

)

f

(

a

)

Note : degree fahrenheit ( x ) and f ( a ) are the like thing, and are just different notations for writing the rule. “ heat content ” placid represents the difference between a or x-values .
This theory holds true for all functions, even though we derived it from a telephone line. consequently, this is the only difference quotient formula you need to know !
When using the difference quotient, several questions types can be asked. Most much, the questions found will be either to set up a difference quotient for a given routine or to simplify f ( a+h ) −f ( a ) h\frac { f ( a+h ) -f ( a ) } { planck’s constant } high frequency ( a+h ) −f ( a ) ​ for each function. fortunately, it is very easy to set up and use, once you are companion with it. It is most difficult to simplify with rational number functions and radical functions, but still more than manageable with lots of practice !
As always, the best room to practice this is to take a expect at some exercise problems. But, before we do that, make certain you ‘re comfortable with some basic operations with functions by checking out our videos on serve notation, dividing functions, composite functions, and the slope equation. Having mastery of all of these techniques will make learning how to use the dispute quotient much much easier !

## How to Find the Difference Quotient:

In the most basic sense, setting up the deviation quotient merely involves subbing in the proper terms into the formula degree fahrenheit ( a+h ) −f ( a ) h\frac { f ( a+h ) -f ( a ) } { hydrogen } high frequency ( a+h ) −f ( a ) ​. however, to make the process easier, it is best to take a estimable search at the given equation to make certain it is in the simplest form possible. Below is an case of what this means :

f(x)=2×2+5x+4f(x) = 2x^{2} + 5x + 4

f

(

x

)

=

2

x

2

+

5

x

+

4

f(x)=3(x−4x)2−3x+4(−8)f(x) = 3(x-4x)^{2} – 3x + 4(-8)

f

(

x

)

=

3

(

x

4

x

)

2

3

x

+

4

(

8

)

As you can see, when trying to sub in terms into the deviation quotient formula, the foremost equation will be much easier than the moment. This is because the second equation involves a bunch more operations, and the addition of “ hydrogen ” into that mix makes errors much more likely. This will become more apparent as we get into some example problems .

## Set Up a Difference Quotient for Linear Functions:

As it often goes, setting up the dispute quotient is easiest for analogue functions. For this kind of work, the less variables and the lower the degree, the easier it is !
Example 1:
Find the remainder quotient of farad ( x ) =2x+5f ( x ) = 2x+5f ( ten ) =2x+5

• STEP 1: Set Up Difference Quotient
• In order to use the difference quotient, all we need to do is set f ( a ) adequate to f ( adam ), and make the necessity modifications to accommodate “ hydrogen ”. It is very important to keep traverse of what is in the parentheses, as this is where it is most common to make mistakes ! Once we do all this, we plainly substitute what we need from f ( x ) to set up the difference quotient .
here ‘s the difference quotient formula again .
fluorine ( a+h ) −f ( a ) h\frac { farad ( a+h ) -f ( a ) } { planck’s constant } high frequency ( a+h ) −f ( a ) ​
Let ‘s find farad ( x+h ) by substituting ( adam + hydrogen ) into farad ( x ) .
fluorine ( x+h ) =2 ( x+h ) +5f ( x+h ) = 2 ( x+h ) +5f ( x+h ) =2 ( x+h ) +5
then, we can plug farad ( ten + planck’s constant ) and f ( ten ) into the formula .
= 2 ( x+h ) +5− ( 2x+5 ) h\frac { 2 ( x+h ) +5- ( 2x+5 ) } { henry } h2 ( x+h ) +5− ( 2x+5 ) ​
• STEP 2: Simplify
once we have set up the remainder quotient, all we need to do is expand whatever is in the brackets, collect like terms, and simplify !
2 ( x+h ) +5− ( 2x+5 ) h\frac { 2 ( x+h ) +5- ( 2x+5 ) } { planck’s constant } h2 ( x+h ) +5− ( 2x+5 ) ​
Expand the brackets, and collect like terms .
2x+2h+5−2x−5h=2hh\frac { 2x+2h+5-2x-5 } { henry } = \frac { 2h } { heat content } h2x+2h+5−2x−5​=h2h​
immediately simplify the fraction by cancelling out henry on the numerator and denominator .
= 2hh=2\frac { 2h } { planck’s constant } = 2h2h​=2
So we know the deviation quotient of fluorine ( x ) =2x+5 is 2

## Set Up a Difference Quotient for Polynomial Functions:

polynomial functions are of like trouble when it comes to finding the difference quotient. Again, as the degree ( the highest exponent ) of the polynomial gets higher, as does the difficulty. The most common difference quotient problems will deal with quadratic polynomial functions that have a degree of 2. late, we will besides look at a higher academic degree polynomial serve by setting up a difference quotient for a cubic serve .
Example 2:
Find the difference quotient of fluorine ( x ) =x2+4x−6f ( x ) = x^ { 2 } +4x-6f ( x ) =x2+4x−6

• STEP 1: Set Up Difference Quotient
• again, in order to use the difference quotient, all we need to do is set f ( a ) adequate to f ( adam ), and make the necessity modifications to accommodate “ hydrogen ”. besides remember it is very significant to keep path of what is in the parentheses, as this is where it is most common to make mistakes ! As an authoritative side note, make indisputable you tackle these kinds of problem when the serve is expanded out amply ! If this function had been in factor form, it would have been much harder to set up the quotient .
fluorine ( a+h ) −f ( a ) h\frac { degree fahrenheit ( a+h ) -f ( a ) } { heat content } high frequency ( a+h ) −f ( a ) ​
= ( x+h ) 2+4 ( x+h ) −6− ( x2+4x−6 ) h\frac { ( x+h ) ^ { 2 } +4 ( x+h ) -6- ( x^ { 2 } +4x-6 ) } { heat content } h ( x+h ) 2+4 ( x+h ) −6− ( x2+4x−6 ) ​
• STEP 2: Simplify
barely like last exemplar, once we have set up the difference quotient, all we need to do is expand whatever is in the brackets, collect like terms, and simplify !
= ( x+h ) 2+4 ( x+h ) −6− ( x2+4x−6 ) h\frac { ( x+h ) ^ { 2 } +4 ( x+h ) -6- ( x^ { 2 } +4x-6 ) } { planck’s constant } planck’s constant ( x+h ) 2+4 ( x+h ) −6− ( x2+4x−6 ) ​
= x2+2xh+h2+4x+4h−6−x2−4x+6h\frac { x^ { 2 } +2xh+h^ { 2 } +4x+4h-6-x^ { 2 } -4x+6 } { heat content } hx2+2xh+h2+4x+4h−6−x2−4x+6​
= 2xh+h2+4hh\frac { 2xh+h^ { 2 } +4h } { planck’s constant } h2xh+h2+4h​
= planck’s constant ( 2x+h+4 ) h\frac { hydrogen ( 2x+h+4 ) } { henry } hh ( 2x+h+4 ) ​
= 2x+h+42x+h+42x+h+4

following, let ‘s take a look at a more difficult polynomial by dealing with a cubic routine .
Example 3:
Find the remainder quotient of farad ( x ) =x3−6×2+11x−6f ( x ) = x^ { 3 } -6x^ { 2 } +11x-6f ( x ) =x3−6×2+11x−6

• STEP 1: Set Up Difference Quotient
• barely like with the quadratic routine in the previous exemplar, in order to use the difference quotient, all we need to do is set f ( a ) equal to f ( x ), and make the necessity modifications to accommodate “ henry ”. besides remember it is very authoritative to keep track of what is in the parentheses, as this is where it is most common to make mistakes ! This can not be more significant as we start dealing with more unmanageable equations and functions. ultimately, just like the end case, make certain you set up the difference quotient for the function in its in full expanded mannequin, i.e. not factor !

Set up difference quotient

• STEP 2: Simplify
once we have set up the difference quotient, all we need to do is expand whatever is in the brackets, collect like terms, and simplify ! This is much more unmanageable compared to the quadratic case, as we have many more terms ! Be sure to double and triple check your function. It is very easy to lose track of terms in these kinds of questions .

Simplify

• Finally, we are left with the final answer of:
= 3×2+3xh+h2−12x−6h+113x^ { 2 } +3xh+h^ { 2 } -12x-6h+113×2+3xh+h2−12x−6h+11

As you can imagine, dealing with tied higher-order polynomials can be quite the challenge. We wo n’t cover anything more difficult than cubic functions in this article, but it would n’t hurt to challenge yourself ! Practice makes perfect .

## Set Up a Difference Quotient for Rational Functions:

Polynomials and linear functions are often the simplest functions to deal with when doing problems on the remainder quotient. The future sections cover rational and radical functions, which involve an extra floor of trouble .
Example 4:
Find the remainder quotient of degree fahrenheit ( x ) =1xf ( x ) = \frac { 1 } { adam } f ( x ) =x1​

• STEP 1: Set Up Difference Quotient
• even though we have a intellectual officiate, which is a moment more crafty, the technique is still the like. In order to use the dispute quotient, all we need to do is set f ( a ) peer to f ( adam ), and make the necessity modifications to accommodate “ hydrogen ”. Since we are using a rational function, you need to pay extra attention to detail to make certain everything is set up correctly .
degree fahrenheit ( a+h ) −f ( a ) h\frac { f ( a+h ) -f ( a ) } { henry } high frequency ( a+h ) −f ( a ) ​
1x+h−1xh\frac { \frac { 1 } { x+h } -\frac { 1 } { ten } } { henry } hx+h1​−x1​​
• STEP 2: Simplify
Keeping in mind the careful attention we need to have, the routine is still the same ! All we need to do is expand whatever is in the brackets, collect like terms, and simplify !
1x+h−1xh\frac { \frac { 1 } { x+h } -\frac { 1 } { ten } } { h } hx+h1​−x1​​
= 1x+h−1xh∙x ( x+h ) ten ( x+h ) \frac { \frac { 1 } { x+h } -\frac { 1 } { adam } } { heat content } \bullet \frac { x ( x+h ) } { ten ( x+h ) } hx+h1​−x1​​∙x ( x+h ) ten ( x+h ) ​
= x− ( x+h ) ten ( x+h ) h\frac { \frac { x- ( x+h ) } { ten ( x+h ) } } { henry } hx ( x+h ) x− ( x+h ) ​​
= x−x−hx ( x+h ) h\frac { \frac { x-x-h } { adam ( x+h ) } } { planck’s constant } hx ( x+h ) x−x−h​​
= −hx ( x+h ) h\frac { \frac { -h } { ten ( x+h ) } } { henry } hx ( x+h ) −h​​
= −1x ( x+h ) \frac { -1 } { ten ( x+h ) } x ( x+h ) −1​

This is the most basic rational routine that exists, so, let ‘s try something a little bite more unmanageable !
Example 5:
Find the difference quotient of degree fahrenheit ( x ) =1x+1f ( x ) = \frac { 1 } { x+1 } degree fahrenheit ( x ) =x+11​

• STEP 1: Set Up Difference Quotient
• once again, even though we have a rational function, which is a moment more catchy, the technique is still the like. In order to use the remainder quotient, all we need to do is set f ( a ) equal to f ( ten ), and make the necessary modifications to accommodate “ h ”. Since we are using a rational number affair, you need to pay extra attention to detail to make certain everything is set up correctly .
In this case, a slenderly remainder method acting is shown below. alternatively of simply trying to input all values into the rule for the deviation quotient, the first step is to alternatively find f ( x+h ) first. Once we know fluorine ( x+h ), can can sub that into the dispute quotient equation .

Set up difference quotient

• STEP 2: Simplify
After the slightly different first step, the respite is the lapp. Keeping in mind the careful attention we need to have, all we need to do is expand whatever is in the brackets, collect like terms, and simplify ! then, we can easily find our final examination answer .

Simplify

## Set Up a Difference Quotient for Radical Functions:

final, but surely not least, is finding the remainder quotient for radical functions. By their very nature, group functions are a piece more unmanageable than anything else. But, if we stay we the like proficiency we ‘ve been using since linear functions, nothing is all that different !
Example 6:
Find the remainder quotient of degree fahrenheit ( x ) =xf ( x ) = \sqrt { ten } fluorine ( x ) =x​

• STEP 1: Set Up Difference Quotient
• precisely like with rational functions, do n’t let the slippery nature of free radical functions fool you. The procedure is the accurate same. In order to use the dispute quotient, all we need to do is set f ( a ) equal to f ( ten ), and make the necessity modifications to accommodate “ planck’s constant ”. besides, since we are using a radical serve, you need to pay extra attention to detail to make surely everything is set up correctly .
degree fahrenheit ( x+h ) −f ( x ) h\frac { degree fahrenheit ( x+h ) -f ( ten ) } { h } high frequency ( x+h ) −f ( x ) ​ = x+h−xh\frac { \sqrt { x+h } – \sqrt { adam } } { heat content } hx+h​−x​​
• STEP 2: Simplify
Paying careful attention to the ancestor signs and all of our terms, all we need to do is expand whatever is in the brackets, collect like terms, and simplify ! then, we can well find our concluding answer.

x+h−xh=x+h−xh∙x+h+xx+h+x\frac { \sqrt { x+h } – \sqrt { adam } } { planck’s constant } = \frac { \sqrt { x+h } – \sqrt { ten } } { hydrogen } \bullet \frac { \sqrt { x+h } + \sqrt { adam } } { \sqrt { x+h } + \sqrt { x } } hx+h​−x​​=hx+h​−x​​∙x+h​+x​x+h​+x​​
hh ( x+h+x ) =1x+h+x\frac { henry } { hydrogen ( \sqrt { x+h } + \sqrt { x } ) } = \frac { 1 } { \sqrt { x+h } +\sqrt { x } } h ( x+h​+x​ ) h​=x+h​+x​1​

And that ‘s all there is to it ! In this article, we have gone over the basic steps to solve any dispute quotient problem. But, your studying is not done there ! Be sure to practice with evening more complex functions to make sure you have a firm sympathy of this topic. To aid in your practice using the difference quotient, check out this great calculator here. full fortune !