Problem 73 - College Algebra with Modeling and Visualization

Question

Complete the following for the given \(f(x)\) (a) Find \(f(x+h)\) (b) Find the difference quotient of \(f\) and simplify. $$ f(x)=2 x+1 $$

Step-by-Step Solution

Answer

The difference quotient for \(f(x) = 2x + 1\) is 2.

Step 1: Write Down Expression for f(x+h)

To find \(f(x+h)\), replace every instance of \(x\) in \(f(x)\) with \(x+h\). Thus, \(f(x) = 2x + 1\) becomes:\[f(x+h) = 2(x+h) + 1\]

Step 2: Simplify f(x+h)

Expand and simplify the expression for \(f(x+h)\):\[f(x+h) = 2(x+h) + 1 = 2x + 2h + 1\]

Step 3: Set Up The Difference Quotient

The difference quotient is given by the formula:\[ \frac{f(x+h) - f(x)}{h} \]Substitute \(f(x+h)\) from Step 2 and \(f(x)\) from the problem statement:\[ \frac{(2x + 2h + 1) - (2x + 1)}{h} \]

Step 4: Simplify The Difference Quotient

Simplify the difference:\[(2x + 2h + 1) - (2x + 1) = 2h\]So, the difference quotient becomes:\[ \frac{2h}{h} \]Simplify further by cancelling the \(h\) terms:\[ 2 \]

Key Concepts

Function Evaluation

When it comes to working with functions, a common task is the evaluation of the function at given points. This involves substituting specific values into the function.

• In this exercise, we are tasked with evaluating the function at \(x+h\), which means replacing each occurrence of \(x\) in the function \(f(x)=2x+1\) with \(x+h\).
• This substitution leads us to the new expression \(f(x+h) = 2(x+h) + 1\).

By performing function evaluation like this, we can explore how changes in the input affect the output, which is a fundamental step in calculus and algebra.

Simplifying Expressions

Simplifying expressions is about making mathematical expressions as efficient and manageable as possible. It's like cleaning up a cluttered room—we're getting rid of what's unnecessary to make calculations clearer and more likely to lead us to a correct answer.

• In the step-by-step solution, after substituting in \(f(x+h) = 2(x+h) +1\), simplification occurs by expanding the expression: \(2(x+h) = 2x + 2h\).
• This expansion leads us to the streamlined \(f(x+h) = 2x + 2h + 1\), which shows each part of the expression in its simplest components without changing its original value.

Simplification is crucial in algebra as it makes dealing with complex expressions easier and helps in understanding the relationships between different parts of an equation.

Algebraic Manipulation

Algebraic manipulation is a fancy way of saying we're moving things around algebraically to see relationships or solve problems. This may involve factoring, expanding expressions, or canceling common terms.

• In this exercise, we set up and simplify the difference quotient: \( \frac{f(x+h) - f(x)}{h}\).
• Substituting the expressions, we find \( \frac{(2x + 2h + 1) - (2x + 1)}{h} \).
  • The algebraic manipulation shows us: \((2x + 2h + 1) - (2x + 1) = 2h\).
  • Which simplifies beautifully to: \ \frac{2h}{h} = 2 \.

This step demonstrates how algebra allows us to see the constant nature of the difference quotient for linear functions. Understanding these manipulations builds the foundation for more advanced calculus concepts like derivatives.