Problem 60 - Intermediate Algebra

Question

Find \(\frac{f(a+h)-f(a)}{h}\) for each of the given functions. (Objective 4) $$f(x)=x^{2}+5$$

Step-by-Step Solution

Answer

\(\frac{f(a+h)-f(a)}{h} = 2a + h\)

Step 1: Identify the function and expression

Given the function is \(f(x) = x^2 + 5\). We need to compute \(\frac{f(a+h)-f(a)}{h}\).

Step 2: Substitute into the expression

Substitute \(a + h\) into the function: \(f(a+h) = (a+h)^2 + 5\). Compute \(f(a)\): \(f(a) = a^2 + 5\).

Step 3: Expand and simplify

Expand \((a+h)^2\) to get \(a^2 + 2ah + h^2\). Therefore, \(f(a+h) = a^2 + 2ah + h^2 + 5\).

Step 4: Calculate the difference

Find \(f(a+h) - f(a)\): \[(a^2 + 2ah + h^2 + 5) - (a^2 + 5) = 2ah + h^2\].

Step 5: Divide by h

Divide the result by \(h\): \[\frac{f(a+h)-f(a)}{h} = \frac{2ah + h^2}{h} = \frac{h(2a + h)}{h}\].

Step 6: Simplify the expression

Cancel \(h\) (since \(h eq 0\)), resulting in \(2a + h\).

Key Concepts

Function Notation

Function notation is a way to represent a function using symbols and letters. Instead of writing out the operation, we use variables to define the function. In this notation, the function is written as \(f(x)\), where "\(f\)" denotes the function, and "\(x\)" is the input variable.
For example, if we have a function \(f(x) = x^2 + 5\), it means that we take any number \(x\), square it, and then add 5.

• The "\(f(x)\)" is read as "\(f\) of \(x\)" and signifies the value of the function at \(x\).
• Function notation allows us to see the relationship between variables in a clear and organized manner.
• This format makes it easier to manage different operations and calculations involving the function.

In exercises involving the difference quotient, we often adjust the input with additional terms like \(a + h\) to find how the output changes. This step is key in calculus for finding rates of change and in understanding the concept of derivatives.

Polynomial Functions

Polynomial functions are expressions consisting of variables raised to whole number powers, combined using addition, subtraction, and multiplication. These are among the most common functions encountered in algebra and calculus.
Take for instance the polynomial \(f(x) = x^2 + 5\). This particular polynomial has:

• A degree of 2, meaning the highest exponent of \(x\) is 2.
• A structure where each term is a product of a constant coefficient and a variable raised to an exponent.

Understanding polynomial functions involves recognizing their degree and knowing how they behave.

• The degree of the polynomial: It tells us the most significant term that influences the overall shape of the graph. Higher degrees indicate more complex curves.
• The constant term: In our example, it is 5, which shifts the entire function up by 5 units when graphed.

Polynomial functions are versatile and can model many real-world situations, from simple equations of motion to complex financial or scientific computations.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and operations. They are the building blocks of algebraic equations.
The expression \(x^2 + 5\) is an example of an algebraic expression. It consists of:

• Two terms: \(x^2\) and 5.
• A variable \(x\) raised to the power of 2 in the term \(x^2\).
• A constant term, which is 5.

Algebraic expressions can be simplified, expanded, or factored based on the rules of algebra.

• Simplifying: Combining like terms or performing operations to reduce the expression to a simpler form.
• Expanding: Multiplying out brackets to present the expression in a longer form.
• Factoring: Rewriting the expression as a product of simplified expressions, which is essential for solving equations.

In exercises involving the difference quotient, simplifying and manipulating algebraic expressions is crucial to isolate and understand changes in functions, particularly when leading towards calculus concepts like limits and derivatives.