Problem 31 - Algebra and Trigonometry

Question

Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0\). $$ f(x)=5 $$

Step-by-Step Solution

Answer

The difference quotient is 0.

Step 1: Evaluate f(a)

Given the function \( f(x) = 5 \), we need to find \( f(a) \). Since \( f(x) = 5 \) is a constant function, \( f(a) = 5 \) regardless of the value of \( a \).

Step 2: Evaluate f(a+h)

Next, evaluate \( f(a+h) \). Again, because the function \( f(x) = 5 \) is constant, \( f(a+h) = 5 \) for any value of \( a \) and \( h \).

Step 3: Calculate the Difference Quotient

Now, compute the difference quotient \( \frac{f(a+h) - f(a)}{h} \). Substitute the values found: \( f(a+h) = 5 \) and \( f(a) = 5 \). This leads to:\[\frac{f(a+h) - f(a)}{h} = \frac{5 - 5}{h} = \frac{0}{h} = 0\]

Step 4: Conclusion

The difference quotient simplifies to 0. The function \( f(x) = 5 \) results in a difference quotient of 0 when evaluated with these values.

Key Concepts

Constant Function

A constant function is one of the simplest types of functions in mathematics. It is defined by the fact that its output value does not change regardless of the input. The general form of a constant function is \( f(x) = c \) where \( c \) is a constant. This means that no matter what value is substituted for \( x \), the result will always be \( c \).
For example, the function \( f(x) = 5 \) is constant because for any input \( x \), the output will always be 5.

• Constant functions graph as horizontal lines in the Cartesian coordinate system.
• They have a slope of 0, indicating no change.
• In real-world applications, constant functions typically represent fixed rates or quantities.

Understanding constant functions is fundamental as they provide a basis for more complex function analysis.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input. It involves replacing the variable in the function with the given input and calculating the resultant value.
For the function \( f(x) = 5 \), evaluating the function at some particular input \( a \) requires substituting \( a \) into the function, simply resulting in 5 because the function is constant.

• Evaluate \( f(a) \): Replace \( x \) with \( a \), giving \( f(a) = 5 \).
• Evaluate \( f(a + h) \): Similarly, replace \( x \) with \( a + h \), leading to \( f(a + h) = 5 \).

Function evaluation clarifies the behavior of functions and how input values influence output results.

Algebraic Expression

An algebraic expression is a mathematical phrase that can include numbers, operations, and variables. Expressions are essential in forming equations and understanding relationships between variables.
In the context of this problem, the example of an algebraic expression can be seen in the calculation of the difference quotient, \[ \frac{f(a+h) - f(a)}{h} \].
This expression determines the average rate of change between two points on a function, an important concept in calculus.

• Difference quotients are particularly useful for analyzing non-constant functions where \( f(a+h) \) differs from \( f(a) \).
• In this specific instance, because \( f(x) = 5 \), the expression simplifies to zero, indicating no change.
• Simplifying algebraic expressions often involves combining like terms or reducing fractions as shown when eliminating the \( h \) in \( \frac{0}{h} \) to result in 0.

Mastering algebraic expressions is critical for solving and simplifying mathematical problems effectively.