Problem 105 - College Algebra Essentials

Question

Explain how to find the difference quotient of a function \(f\) \(\frac{f(x+h)-f(x)}{h},\) if an equation for \(f\) is given.

Step-by-Step Solution

Answer

The difference quotient of a function \(f(x)\) is computed using the formula \(\frac{f(x+h)-f(x)}{h}\). This involves substituting \(x+h\) to the function to obtain \(f(x+h)\), subtracting \(f(x)\) from \(f(x+h)\) and dividing the result by \(h\). Make sure to fully simplify the result if possible.

Step 1: Identify the function \(f(x)\)

Given a function \(f(x)\), the first step is to identify it. The function could be any mathematical expression in terms of \(x\). \(f(x)\) could be a linear function like \(2x + 3\), a quadratic function such as \(x^2 + 3x + 2\), or any other polynomial functions, or may even involve trigonometric, exponential, logarithmic functions etc.

Step 2: Compute \(f(x+h)\)

This step involves substituting \(x+h\) into the function \(f(x)\) to get a new function \(f(x+h)\). This could involve expanding brackets for polynomial functions or using trigonometric identities for trigonometric functions, for example.

Step 3: Subtract \(f(x)\) from \(f(x+h)\)

Next, subtract the function \(f(x)\) from the newly obtained function \(f(x+h)\). This will form the numerator of the difference quotient. Make sure to enclose \(f(x)\) within brackets while subtracting. This is because \(f(x)\) could be a sum or difference of several terms, thus failing to do so might lead to errors in signs.

Step 4: Divide by \(h\)

Divide the result of Step 3, which forms the numerator, by \(h\) to compute the difference quotient. This would sometimes lead to a more complex fraction. However, ensure that all terms and factors are fully simplified. Try to cancel out any common factors from the numerator and denominator if possible. Although not always necessary, it provides a less complex and more standard form of the result. If the function was polynomial or any other simpler functions, \(h\) in the denominator could often cancel out, leaving an expression that does not involve \(h\).