Problem 37 - 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 function values are both 5, and the difference quotient is 0.

Step 1: Understanding the Function

The given function is \( f(x) = 5 \). This means that for any input \( x \), the output is always \( 5 \). It is a constant function.

Step 2: Calculate \( f(a) \)

Since \( f(x) = 5 \) for any \( x \), substituting \( a \) into the function gives \( f(a) = 5 \).

Step 3: Calculate \( f(a+h) \)

Similarly, since \( f(x) = 5 \), substituting \( a+h \) into the function gives \( f(a+h) = 5 \).

Step 4: Find the Difference Quotient

The difference quotient is given by \( \frac{f(a+h)-f(a)}{h} \). Substituting the previously found values, we get:\[\frac{5 - 5}{h} = \frac{0}{h} = 0.\] Since the numerator is zero, the difference quotient simplifies to \( 0 \), regardless of the value of \( h \) (as long as \( h eq 0 \)).

Key Concepts

Constant Function

A constant function is a type of function where the value of the output is the same, no matter what input you choose. This means, if you're working with a constant function like the one in this exercise, which is \( f(x) = 5 \), the output will always be 5 for any value of \( x \). This characteristic makes constant functions very easy to work with, as you don’t have to worry about changing outputs; they stay consistent.
Here are some key points to remember about constant functions:

• Their graph is a horizontal line on the Cartesian plane.
• They have a slope of zero because there is no change in the output value.
• In terms of algebra, their equation typically looks like \( f(x) = c \), where \( c \) is a constant value.

In our exercise, this consistency leads directly to simple calculations for \( f(a) \) and \( f(a+h) \), both consistently resulting in 5.

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input value. In simpler terms, it means "plugging in" a number into the function to see what result you get. In our given problem, we have a constant function \( f(x) = 5 \). Let's break down what function evaluation entails:

• When you evaluate \( f(a) \), you're asking: "What is the value of the function when \( x \) is \( a \)?" Given \( f(x) = 5 \), it means \( f(a) = 5 \).
• Similarly, \( f(a+h) \) is evaluated by substituting \( a+h \) into the function. Since the function is constant, \( f(a+h) = 5 \) as well.

These evaluative steps might seem trivial in the context of a constant function, but they lay the groundwork for understanding how any function can be analyzed. With other types of functions, such as linear or quadratic functions, the process remains vital for determining outputs, even though the results will differ.

Algebraic Functions

Algebraic functions are composed of algebraic expressions, which include operations like addition, subtraction, multiplication, division, and powers with whole number exponents. The function \( f(x) = 5 \) is an example of a simple algebraic function. Although constant, it still falls under this broad category due to its simplicity and structure
Some features of algebraic functions include:

• They can be simple like constant functions or more complex involving polynomials.
• They play a crucial role in many areas of mathematics including calculus, where understanding their behavior under different operations, such as differentiation, is key.
• They allow for the exploration of limits and the behavior of functions as variables approach certain values.

In our exercise, the difference quotient \( \frac{f(a+h)-f(a)}{h} \) evaluates to zero because both \( f(a+h) \) and \( f(a) \) give the same constant value, clearly illustrating how algebraic manipulation yields valuable insights. While constants make this straightforward, the approach is applicable to more complex algebraic functions.