What a Difference Quotient Measures
A function assigns an output to each input in its domain. A difference quotient compares the change in output with a nonzero change in input. It is an algebraic expression: simplifying it means rewriting that expression in an equivalent form wherever the original expression is defined.
There are two common forms. One uses a variable input \(x\) and an increment \(h\); the other compares a variable input \(x\) with a fixed input \(a\). In either form, the denominator must be nonzero. In this tutorial, the goal is to simplify the expressions by expanding or factoring—not to evaluate a limit or a derivative.
The numerator records the difference between two function values. The denominator records the corresponding difference between the inputs. For algebraic simplification, the key task is often to show that the numerator contains the denominator as a factor. Cancellation is valid only where that factor is nonzero, so the conditions in the definition remain important.
Expanding and Factoring with an Increment
For a polynomial, substituting \(x+h\) for \(x\) typically creates terms involving \(h\). Subtracting \(f(x)\) cancels some terms. The remaining numerator often has a factor of \(h\), which can then be cancelled against the denominator, because the difference quotient already requires \(h\ne0\).
Worked Example: A Quadratic with a Linear Term
Let \(f(x)=x^2+3x-2\). Simplify \(\frac{f(x+h)-f(x)}{h}\), where \(h\ne0\).
First substitute \(x+h\) into the function:
Now subtract \(f(x)=x^2+3x-2\). Keep the parentheses so each term in \(f(x)\) is subtracted:
Therefore, since \(h\ne0\), cancel the shared factor:
Check by multiplying the simplified expression by \(h\): \(h(2x+h+3)=2xh+h^2+3h\), exactly the numerator obtained before cancellation. The simplification is valid for every \(x\) and every nonzero \(h\).
The cancellation does not mean that the original quotient is defined at \(h=0\). The original expression has denominator zero there. The simplified expression \(2x+h+3\) has a value at \(h=0\), but that fact does not change the domain of the original quotient.
A Quotient from a Fixed Input
When the second input is a fixed number \(a\), it can be efficient to evaluate \(f(a)\) first. Then factor the numerator \(f(x)-f(a)\). If the factor \(x-a\) appears, it can be cancelled for \(x\ne a\). This is the same algebraic principle as cancelling \(h\) in an increment quotient, with \(x-a\) playing the role of the nonzero input change.
Worked Example: Factoring Around a Fixed Input
Let \(f(x)=x^2-4x+1\), and simplify \(\frac{f(x)-f(2)}{x-2}\) for \(x\ne2\).
Evaluate the function at the fixed input:
Substitute this value into the numerator, then factor:
Since \(x\ne2\), cancel one factor of \(x-2\):
Check at an allowed input, say \(x=3\). The original quotient is \(\frac{f(3)-f(2)}{3-2}\). Since \(f(3)=9-12+1=-2\) and \(f(2)=-3\), it equals \(\frac{-2-(-3)}{1}=1\). The simplified expression gives \(3-2=1\). Also, multiplying the simplified expression by \(x-2\) gives \((x-2)^2=x^2-4x+4\), the original numerator. The original quotient still excludes \(x=2\).
Why Factoring Produces the Simplified Expression
The algebra behind the method can be seen for any quadratic polynomial \(f(x)=Ax^2+Bx+C\), where \(A\), \(B\), and \(C\) are constants. For a nonzero increment \(h\), substitute \(x+h\), subtract \(f(x)\), and collect terms:
Thus, wherever \(h\ne0\), the difference quotient has the form
This derivation justifies the cancellation for every quadratic, rather than only for one example: expansion shows that every term remaining in the numerator contains \(h\). The constant term \(C\) cancels, and the terms without \(h\) cancel as well. The derivation concerns only an algebraic identity on the stated domain; it does not require or evaluate a limit.
Difference Quotients with Rational Functions
The same steps apply when a function has a restricted domain. Before simplifying, check that each function value in the quotient is defined, as well as checking that the denominator of the quotient is nonzero. Cancelling a factor can make the final expression look defined at additional inputs, but it does not add those inputs to the domain of the original quotient.
Worked Example: Keeping Track of Domain Restrictions
Let \(f(x)=\frac{1}{x}\). Simplify \(\frac{f(x+h)-f(x)}{h}\).
For the two function values to be defined, we need \(x\ne0\) and \(x+h\ne0\). The difference quotient also requires \(h\ne0\). Combine the two fractions in the numerator:
The factor \(h\) cancels because \(h\ne0\). The simplified expression is valid under all the original conditions: \(x\ne0\), \(x+h\ne0\), and \(h\ne0\). To check, multiply the final expression by \(h\): \(h\left(-\frac{1}{x(x+h)}\right)=-\frac{h}{x(x+h)}\), the combined numerator difference. For example, with \(x=2\) and \(h=1\), the original quotient is \(\frac{\frac13-\frac12}{1}=-\frac16\); the simplified expression is \(-\frac{1}{2(3)}=-\frac16\).
A Common Error: Losing a Term When Expanding
When expanding a square, the middle term is essential. A frequent mistake is to write \((x+h)^2=x^2+h^2\), omitting the product term \(2xh\). That changes the numerator and produces an incorrect quotient.
Check the difference numerically at \(x=1\), \(h=1\). The original quotient is \(\frac{f(2)-f(1)}{1}=\frac{4-1}{1}=3\). The incorrect result gives \(h=1\), while the corrected expression gives \(2(1)+1=3\). A numerical check can expose an error, while the expansion establishes the correct identity for all allowed \(x\) and \(h\).
A Reliable Simplification Process
For either form of difference quotient, keeping the domain conditions visible helps prevent invalid cancellation. A useful sequence is:
Require a nonzero denominator and require every function input to belong to the function's domain.
Write \(f(x+h)\) or \(f(a)\) using parentheses, so the entire input is substituted.
Expand polynomial terms and combine like terms, or use a common denominator when the numerator contains fractions.
Rewrite the numerator to display a factor matching the denominator.
Multiply the simplified expression by the cancelled factor to recover the numerator, and retain the original restrictions.
The important distinction is between an expression and its domain. If \(h\ne0\), then \(\frac{h(2x+h+3)}{h}\) equals \(2x+h+3\). The equality is valid under the nonzero condition, even though the latter expression can be evaluated at \(h=0\). Cancellation simplifies values on the original domain; it does not erase the condition that made the cancellation valid.
Check Your Understanding
For each question, simplify only where the original quotient is defined. Keep track of any excluded inputs.
- For \(f(x)=x^2+2x\), expand and simplify \(\frac{f(x+h)-f(x)}{h}\). State the condition on \(h\).
- For \(f(x)=x^2+1\), simplify \(\frac{f(x)-f(3)}{x-3}\). Which input is excluded?
- For \(f(x)=\frac{1}{x}\), list all restrictions on \(x\) and \(h\) in \(\frac{f(x+h)-f(x)}{h}\) before cancelling.
- Why is \((x+h)^2\) not equal to \(x^2+h^2\)? Write the correct expansion.
- After cancelling \(h\) from a difference quotient, why must the original condition \(h\ne0\) still be retained?
