The “difference quotient” taken at x for a function f is defined as: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$ , with $$ h \ne 0 $$ ’s take a look at a graphical representation.
Assume that the black graph below is a graph of some function f. Then, you will notice that there are two points highlighted in yellow on the graph: $$ \left( {x,f\left( x \right)} \right) $$ and $$ \left( {x + h,f\left( {x + h} \right)} \right) $$ . Notice that h is just the distance added to the first input, x, to obtain the second input, x + h. A red line is drawn through these points. This line, because it goes through two points on the curve, is called a secant line. What is the slope of this secant line?
Recall that $$ slope = {\Large{{\Delta y} \over {\Delta x}}} $$. So, here, the slope is: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over {\left( {x + h} \right) - x}}} = {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$. Notice that this expression is the difference quotient. So, in essence, a difference quotient is the slope of a secant line of the graph of a function.
Why do we care to find the slope of a secant line? We care mainly because it is a fair approximation for the slope of the curve between those points. In fact, the smaller we make h (which, again, is the distance between the input values of the two points), the better approximation we have for the slope of a small part of the curve. This becomes very important, as you will find later in your study of calculus.
For now, we will practice calculating the difference quotient for some different functions.
Let $$
f\left( x \right) = 3x - 7
$$
. Find and simplify the difference quotient for f.
First, recall that the difference quotient is:
$$
{\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}}
$$, with $$
h \ne 0
$$
So, first, we want to find $$ f\left( {x + h} \right) $$ . That is, we want to find the value of the function when we input $$ x + h $$ Inputs and Outputs of Functions).
In this case:
$$
f\left( {x + h} \right) = 3\left( {x + h} \right) - 7 = 3x + 3h - 7
$$
We were given $$ f\left( x \right) = 3x - 7 $$
$$
{\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h} = {{\left( {3x + 3h - 7} \right) - \left( {3x - 7} \right)} \over h}}
$$
Now, use the distributive property to distribute the negative sign:
$$
{\Large{{\left( {3x + 3h - 7} \right) - \left( {3x - 7} \right)} \over h} = {{3x + 3h - 7 - 3x + 7} \over h}}
$$
Then, combine like terms in the numerator:
$$
{\Large{{3x + 3h - 7 - 3x + 7} \over h} = {{3h} \over h}}
$$
Finally, cancel the h’s, and we get that the difference quotient is:
$$
{\Large{{3h} \over h}} = 3
$$
So, this means that, for this function, the slope of the secant line for any value of h and at any x is going to be 3. This makes sense because$$ f\left( x \right) = 3x - 7 is a linear function with a slope of 3.
Let $$
f\left( x \right) = 3x^2 - 4x + 5
$$
. Find the difference quotient for f.
First, let’s find $$
f\left( {x + h} \right)
$$ (again, for help on this, see Inputs and Outputs of Functions).
$$
f\left( {x + h} \right) = 3\left( {x + h} \right)^2 - 4\left( {x + h} \right) + 5
$$
Now, we need to multiply polynomials in order to square $$ x + h $$
$$
f\left( {x + h} \right) = 3\left( {x + h} \right)^2 - 4\left( {x + h} \right) + 5 = 3\left( {x^2 + 2xh + h^2 } \right) - 4\left({x + h} \right) + 5
$$
Using the distributive property, we get:
$$
f\left( {x + h} \right) = 3\left( {x^2 + 2xh + h^2 } \right) - 4\left( {x + h} \right) + 5 = 3x^2 + 6xh + 3h^2 - 4x - 4h + 5$$
Now, we can put $$ f\left( {x + h} \right) $$ and our given $$ f\left( x \right) into the difference quotient formula to get:
$$
{\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h} = {{\left( {3x^2 + 6xh + 3h^2 - 4x - 4h + 5} \right) - \left({3x^2 - 4x + 5} \right)} \over h}}
$$
Distributing the negative, we get:
$$
{\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h} = {{3x^2 + 6xh + 3h^2 - 4x - 4h + 5 - 3x^2 + 4x - 5} \over h}}
$$
Combining like terms, we get:
$$
{\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h} = {{6xh + 3h^2 - 4h} \over h}}
$$
We can now factor an h out of the numerator, which cancels with the denominator, to get:
$${\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} = {\Large{{h\left( {6x + 3h - 4} \right)} \over h}} = 6x + 3h - 4$$.
So, this time, you can see that our difference quotient depends on the x for which you would like to determine it, as well as the size of h.
Some practice problems to check your skills:
1. Let $$ f\left( x \right) = 17 $$ . Find and simplify the difference quotient: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$
Think about it for a moment and then access this link 
to view answer.
2. Let $$ f\left( x \right) = 4x + 15 $$ . Find and simplify the difference quotient: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$
Think about it for a moment and then access this link to view answer.
3. Let $$ f\left( x \right) = - 5x^2 - 3x + 9 $$ . Find and simplify the difference quotient: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$
Think about it for a moment and then access this link to view answer.
4. Let $$ f\left( x \right) = 4x^3 - 7 $$ . Find and simplify the difference quotient: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$
Think about it for a moment and then access this link to view answer.
5. Let $$ f\left( x \right) = {\Large{1 \over x}} $$ . Find and simplify the difference quotient: $$ {\Large{{f\left( {x + h} \right) - f\left( x \right)} \over h}} $$
Think about it for a moment and then access this link to view answer.