Problem 69 - College Algebra

Question

Find and simplify the difference quotient $$\frac{f(x+h)-f(x)}{h}, h \neq 0$$ for the given function. $$f(x)=-2 x^{2}-x+3$$

Step-by-Step Solution

Answer

The difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the given function is simplified to, \( -4x -2 -h \).

Step 1: Substitution of \(x+h\) into \(f(x)\)

Substitute \(x+h\) into the function \(f(x)= -2x^{2}-x+3\). This will give \(f(x+h)=-2(x+h)^{2}-(x+h)+3\). Apply the formula for the square of a binomial \((a+b)^2=a^2+2ab+b^2\) and simplify the result.

Step 2: Substitution of \(x\) into \(f(x)\)

Substitute \(x\) into the function \(f(x)= -2x^{2}-x+3\). The function remains the same since we are substituting with an original variable, so \(f(x)=-2x^{2}-x+3\).

Step 3: Computing the Difference Quotient

To compute the difference quotient \(\frac{f(x+h)-f(x)}{h}\), replace \(f(x+h)\) and \(f(x)\) with their respective expressions obtained from the previous steps. Then subtract these two fractions and divide the result by \(h\). Simplify the resulting expression by canceling out any like terms in the numerator and the denominator.

Step 4: Final Simplification

A potentially dense algebra step, but necessary to put the difference quotient in the simplest terms possible. Look to cancel any like terms or to represent the expression in the more simple possible way. The final result will be the difference quotient completely simplified.

Key Concepts

Function Substitution

Function substitution involves replacing variables within a function with new expressions or numbers. Let's look at the function given: \[ f(x) = -2x^2 - x + 3 \] With function substitution, we replace every instance of the variable \(x\) in the function with \(x+h\). This transforms the function into \(f(x+h)\). It's crucial to apply this step carefully in the difference quotient calculation:

• Substitute \(x+h\) into the given polynomial function.
• This results in: \(-2(x+h)^2 - (x+h) + 3\).

Function substitution is foundational in calculus, particularly when evaluating limit processes such as the difference quotient. It effectively shifts the function along the x-axis, helping to analyze how the function behaves as it approaches specific values.

Algebraic Simplification

After substituting \(x+h\) into the function, the expression becomes a bit complex. This is where algebraic simplification comes in handy. Simplification involves using algebraic rules to make expressions easier to interpret and work with. For the expression \(-2(x+h)^2 - (x+h) + 3\), perform the following steps:

• Apply the formula for expanding a binomial: \((a+b)^2 = a^2 + 2ab + b^2\).
• Distribute \(-2\) across each term within the bracket.
• Combine like terms to simplify further.

A neat algebraic simplification enhances clarity, easing calculations in subsequent steps. This process allows us to identify patterns or factors that can be canceled out, thereby simplifying complex expressions.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables, multiplied by coefficients. They are characterized by their degree, determined by the highest power of the variable.In our example, \(f(x) = -2x^2 - x + 3\), the function is a quadratic polynomial because the highest power of \(x\) is 2. Polynomial functions exhibit properties such as:

• Having a smooth and continuous graph.
• Being predictable in their behavior at extreme values of \(x\).

Understanding polynomial functions is critical when dealing with calculus topics like differentiation and integration, as these operations often involve polynomial expressions. Recognizing the type of polynomial helps in applying the correct simplification and differentiation techniques.

Binomial Expansion

The binomial expansion technique is essential when working with expressions like \((x+h)^2\) during function substitution. It allows for the expansion of powers of a binomial using the formula:\[(a+b)^2 = a^2 + 2ab + b^2\]In the context of the difference quotient problem, replacing \((x+h)^2\) with its expanded form, \(x^2 + 2xh + h^2\), simplifies the expression further when calculating the difference quotient.This breakdown using binomial expansion is instrumental in unraveling complexities within algebra. Mastery of this technique ensures precise and easy evaluation and simplification of expressions required in calculus applications.