Problem 117 - College Algebra

Question

Determine whether each statement makes sense or does not make sense, and explain your reasoning. I noticed that the difference quotient is always zero if \(f(x)=c,\) where \(c\) is any constant.

Step-by-Step Solution

Answer

The statement makes sense. As the difference quotient for a given function is a measure of the average rate of change for that function, if the function \(f(x) = c\) is a constant, then the output \(f(x)\) is not changing regardless of the input \(x\). Hence, the difference quotient is zero.

Step 1: Understanding the Statement

The statement says that the difference quotient of a function equals to zero, when the function is a constant. The difference quotient is a measure of the average rate of change of a function over a small interval, and it is defined as \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\] for a function \(f(x)\). A constant function is a function where the output (or the value of \(f(x)\)) is the same no matter what the input \(x\) is.

Step 2: Considering the Constant Function

A function \(f(x) = c\) is a constant function, where \(c\) does not depend on \(x\). For these functions, \(f(x+h) = c\) and \(f(x) = c\) no matter what the values of \(x\) and \(h\) are. This is because the function does not depend on \(x\), hence the outcomes stay the same even when we change \(x\).

Step 3: Computing the Difference Quotient

Substituting \(f(x+h) = c\) and \(f(x) = c\) into the definition of the difference quotient, we get \[f'(x) = \lim_{h \to 0} \frac{c - c}{h}\]. The numerator is \(0\), and dividing \(0\) by any \(h\) brings the result zero. Thus the difference quotient of a constant function \(f(x) = c\) is indeed \(0\), so the statement makes sense.