Problem 71 - 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. $$ f(x)=-2 x $$

Step-by-Step Solution

Answer

(a) \(f(x+h) = -2x - 2h\); (b) Simplified difference quotient: \(-2\).

Step 1: Substitute to find f(x+h)

Start with the function given, \(f(x) = -2x\). To find \(f(x+h)\), replace every instance of \(x\) in the function with \(x+h\): \[f(x+h) = -2(x + h) \] Distribute the \(-2\) through the parentheses: \[ f(x+h) = -2x - 2h \]

Step 2: Set up the difference quotient

The difference quotient is given by the expression: \[\frac{f(x+h) - f(x)}{h}\]Insert \(f(x+h)\) and \(f(x)\) into the difference quotient: \[\frac{(-2x - 2h) - (-2x)}{h}\]

Step 3: Simplify the numerator

Simplify the expression in the numerator by eliminating similar terms: \[(-2x - 2h) - (-2x) = -2x - 2h + 2x \] Combine the \(-2x\) and \(+2x\) terms: \[-2h\] So the numerator is simplified to \(-2h\).

Step 4: Simplify the difference quotient

Substitute the simplified numerator into the difference quotient expression: \[\frac{-2h}{h}\] Cancel the \(h\) terms: \[-2\] Thus, the simplified form of the difference quotient is \(-2\).

Key Concepts

Function Substitution

When you're given a function, it's possible to alter it by substituting within the function itself. This allows us to explore how the function behaves when its inputs change. For the given exercise, we started with the function:

• \(f(x) = -2x\)

To find \(f(x+h)\), substitute \(x+h\) for every \(x\) in the function. This step transforms the original function into:

• \(f(x+h) = -2(x + h)\)

Next, you distribute the \(-2\) across the terms in the parentheses to obtain:

• \(-2x - 2h\)

This substitution tells us how the function changes when \(h\) is added to the input \(x\). It lays the groundwork for the next steps involving the difference quotient.

Algebraic Simplification

Algebraic simplification is crucial when working with difference quotients. Once you've made your substitutions, simplifying the algebraic expressions can make complex calculations much easier. In our example, we found \(f(x+h) = -2x - 2h\) and needed to set up the difference quotient:

• \(\frac{f(x+h) - f(x)}{h}\)

This means we're finding the difference between \(f(x+h)\) and \(f(x)\), and dividing by \(h\):

• \(\frac{(-2x - 2h) - (-2x)}{h}\)

During simplification, it is essential to handle algebraic terms carefully. Particularly, when you subtract terms like \(-2x\), be careful to distribute the negative sign. This will typically involve combining like terms, as follows:

• \(-2x - 2h + 2x\)

After canceling out terms that have identical coefficients, the expression is simplified to \(-2h\). Simplification helps in making the expressions less complex and is vital for finding clear solutions.

Numerator Simplification

In the difference quotient, simplifying the numerator is one of the crucial steps to obtaining a manageable expression. Once you have the expression:

• \((-2x - 2h) - (-2x)\)

Focus on simplifying the numerator by eliminating or combining like terms. In this scenario, combine \(-2x\) and \(+2x\) which results in:

• \(-2h\)

This simplification removes the \(x\) terms from the numerator, leaving only \(-2h\). Next, you substitute this simplified numerator back into the difference quotient:

• \(\frac{-2h}{h}\)

To simplify this expression further, divide both the numerator and the denominator by \(h\), resulting in \(-2\).Numerator simplification is essential because it helps to alleviate unnecessary complexity in calculations by removing common terms. This simplification ultimately leads us to the final simplified form of the difference quotient.