Question

Find the difference quotient of \(f\); that is, find \(\frac{f(x+h)-f(x)}{h}, h \neq 0,\) for each function. Be sure to simplify. \(f(x)=x^{2}-x+4\)

Short answer

\(2x + h - 1\).

Step-by-step solution

- Substitute f(x) and f(x+h) into the difference quotient

First, identify the given function: \(f(x) = x^2 - x + 4\)To calculate the difference quotient \(\frac{f(x+h)-f(x)}{h}\), start by finding \(f(x+h)\).

- Compute f(x+h)

Substitute \(x + h\) into the function \(f(x)\):\(f(x+h) = (x + h)^2 - (x + h) + 4\).Simplify the expression:\(f(x+h) = (x^2 + 2xh + h^2) - x - h + 4 = x^2 + 2xh + h^2 - x - h + 4\).

- Find f(x+h) - f(x)

Now, subtract \(f(x)\) from \(f(x+h)\):\(f(x+h) - f(x) = (x^2 + 2xh + h^2 - x - h + 4) - (x^2 - x + 4)\).Distribute and combine like terms:\(f(x+h) - f(x) = x^2 + 2xh + h^2 - x - h + 4 - x^2 + x - 4 \).This simplifies to:\(f(x+h) - f(x) = 2xh + h^2 - h\).

- Divide by h

Now divide by \(h\), as required in the difference quotient:\(\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - h}{h} \).Factor out the common \(h\) in the numerator:\(\frac{f(x+h) - f(x)}{h} = \frac{h(2x + h - 1)}{h} \).Cancel the \(h\) term:\(\frac{f(x+h) - f(x)}{h} = 2x + h - 1\).

- State the final simplified form

After simplifying, the difference quotient is:\(2x + h - 1\).

Key concepts

Algebra

Algebra is a branch of mathematics that uses symbols and letters to represent numbers and quantities in equations and expressions. Understanding algebra is crucial for solving difference quotient problems.
In this exercise, we deal with polynomials (which are algebraic expressions that involve sums of powers of variables). Specifically, the function given is a quadratic polynomial:

• Given function: \(f(x) = x^2 - x + 4\)

To solve for the difference quotient, we use algebraic techniques such as:

• Substituting variables
• Expanding expressions
• Combining like terms
• Factoring

Using these skills, you can manipulate and simplify expressions to find the difference quotient of a function effectively.

Functions

Functions are mathematical constructs that assign each input exactly one output. They are often written as \(f(x)\), where \(x\) is the input and \(f(x)\) is the output. In this exercise, we have the given quadratic function:

• \(f(x) = x^2 - x + 4\)

To compute the difference quotient, we need to understand the behavior of the function when the input changes slightly. This involves substituting \(x + h\) into the function and finding the new output:

• \(f(x+h) = (x + h)^2 - (x + h) + 4\)

It's important to expand and simplify this expression to see how the function varies with small changes in \(x\). This involves calculating \(f(x + h)\), then subtracting \(f(x)\) and dividing by \(h\):

• \(f(x+h) - f(x)\)
• \(\frac{f(x+h) - f(x)}{h}\)

Understanding how to manipulate functions algebraically is key to mastering the difference quotient.

Simplification

Simplification is the process of reducing an expression to its lowest terms. This is crucial when dealing with complex algebraic expressions. In the difference quotient, simplification makes the final answer easier to work with and understand.
First, express \(f(x + h)\) by substituting and expanding the terms:

• \(f(x + h) = (x + h)^2 - (x + h) + 4\)
• \(f(x + h) = x^2 + 2xh + h^2 - x - h + 4\)

Then, subtract \(f(x)\) from \(f(x+h)\):

• \(f(x+h) - f(x) = x^2 + 2xh + h^2 - x - h + 4 - (x^2 - x + 4)\)
• Combine like terms to simplify: \(2xh + h^2 - h\)

Finally, divide by \(h\) and simplify the expression:

• \(\frac{2xh + h^2 - h}{h} = 2x + h - 1\)

Factoring and simplification techniques are essential to simplify complex algebraic expressions and find the final, simplified form of the difference quotient.