Problem 30 - Algebra and Trigonometry

Question

Find \(f(a), f(a+h),\) and the difference quotient \(\frac{f(a+h)-f(a)}{h},\) where \(h \neq 0\). $$ f(x)=x^{2}+1 $$

Step-by-Step Solution

Answer

\(f(a) = a^2 + 1\), \(f(a+h) = a^2 + 2ah + h^2 + 1\), Difference quotient: \(2a + h\).

Step 1: Find \(f(a)\)

To find \(f(a)\), substitute \(a\) in place of \(x\) in the function formula. The function is \(f(x) = x^2 + 1\). Thus, \(f(a) = a^2 + 1\).

Step 2: Find \(f(a+h)\)

To find \(f(a+h)\), substitute \(a+h\) in place of \(x\) in the function formula. Thus, \(f(a+h) = (a+h)^2 + 1\). Expand \((a+h)^2\) to get \(a^2 + 2ah + h^2\). Thus, \(f(a+h) = a^2 + 2ah + h^2 + 1\).

Step 3: Find the Difference Quotient \(\frac{f(a+h) - f(a)}{h}\)

Substitute \(f(a+h)\) and \(f(a)\) into the difference quotient formula. The difference quotient is \(\frac{f(a+h) - f(a)}{h}\). Substitute the expressions: \(\frac{(a^2 + 2ah + h^2 + 1) - (a^2 + 1)}{h}\). Simplify the numerator: \(a^2 + 2ah + h^2 + 1 - a^2 - 1 = 2ah + h^2\). The difference quotient simplifies to \(\frac{2ah + h^2}{h}\). Factor \(h\) out of the numerator: \(\frac{h(2a + h)}{h}\). Cancel \(h\) to get \(2a + h\).

Key Concepts

Quadratic Functions

Quadratic functions are a special type of polynomial functions characterized by having a degree of 2. This simply means that the highest exponent of the variable in the equation is 2. The general form of a quadratic function is expressed as:

• \( f(x) = ax^2 + bx + c \)

Where:

• \( a, b, \) and \( c \) are constants
• \( a eq 0 \) because if \( a = 0 \), the function is no longer quadratic but becomes linear
• \( x \) represents the variable

This function typically plots as a parabola on a graph, which can open upwards or downwards depending on the sign of \( a \). All quadratic functions share the ability to express relationships where one variable depends on the square of another.

Function Evaluation

Evaluating a function involves substituting a given value into the function and calculating the result. When evaluating a function like \( f(x) = x^2 + 1 \), we replace \( x \) with the given value.

• For instance, to find \( f(a) \), replace \( x \) with \( a \) which gives \( a^2 + 1 \).
• To find \( f(a+h) \), replace \( x \) with \( a+h \). This results in \( (a+h)^2 + 1 \), which further expands to \( a^2 + 2ah + h^2 + 1 \).

This process illustrates how input values are reflected in outputs after substituting them into the function. Understanding this concept makes solving real-world problems easier because it helps us determine outputs based on different input scenarios.

Algebraic Expressions

Algebraic expressions represent mathematical phrases combining numbers, variables, and operations like addition or multiplication. An example is the expression \( (a+h)^2 + 1 \). To manipulate them, you often simplify or expand parts of the expressions to make calculations more manageable.
When dealing with expressions like \( (a+h)^2 \), it's crucial to follow algebraic rules:

• Expand \( (a+h)^2 \) to \( a^2 + 2ah + h^2 \).
• Simplify \( (a^2 + 2ah + h^2 + 1) - (a^2 + 1) \) to \( 2ah + h^2 \).
• Factor to \( h(2a + h) \).
• The presence of like terms means you can combine them using addition or subtraction.

Mastering algebraic expressions is useful for solving complex problems as it allows us to simplify calculations and manipulate equations effectively.