Question

Simplify the difference quotient\(\frac{f(x+h)-f(x)}{h}\) for the following functions. $$f(x)=x^{2}$$

Short answer

Answer: The simplified difference quotient of the given function is \(2x + h\).

Step-by-step solution

Write down the given function and the difference quotient

The given function is \(f(x) = x^2\). The difference quotient is given by \(\frac{f(x+h)-f(x)}{h}\).

Substitute the function into the difference quotient

We will substitute \(f(x)\) and \(f(x+h)\) into the difference quotient formula: $$\frac{f(x+h)-f(x)}{h} = \frac{(x+h)^{2}-x^{2}}{h}.$$

Expand the expression and simplify

Now we will expand and simplify the expression: $$\begin{aligned} \frac{(x+h)^{2}-x^{2}}{h} &= \frac{(x^2 + 2xh + h^2) - x^2}{h} \\ &= \frac{2xh + h^2}{h}. \end{aligned}$$

Factor out the common term "h"

Now we will factor out the common term "h" from the numerator: $$\frac{2xh + h^2}{h} = h(2x + h) \cdot \frac{1}{h}.$$

Cancel out the common term "h"

Since "h" appears both in the numerator and the denominator, we can cancel it out: $$h(2x + h) \cdot \frac{1}{h} = 2x + h.$$

Write the final answer

Therefore, the simplified difference quotient of the given function is \(2x + h\).

Key concepts

Simplification

The process of simplification involves reducing a mathematical expression to its most basic form. When solving any problem, simplifying is often key to making the terms easier to understand and work with. In our exercise, we are focusing on simplifying the difference quotient

• Start by writing down the expression \( \frac{f(x+h)-f(x)}{h} \)
• For \( f(x)=x^2 \), the expression becomes \( \frac{(x+h)^2-x^2}{h} \)

This transformation is essential. It prepares the expression for the next logical steps, which are function substitution and expansion. Simplification often involves combining terms or performing operations like subtraction or division to break down complex expressions.
By making expressions simpler, calculations become more manageable and errors less likely.

Function Substitution

Function substitution is an important skill in mathematics. It involves replacing a variable in a formula with another expression. In our example, we substituted the function \( f(x) \) and \( f(x+h) \) into the difference quotient.

• The original function is \( f(x) = x^2 \)
• Then, replace \( f(x+h) \) as \( (x+h)^2 \)

Doing this substitution, we got the expression \( \frac{(x+h)^2 - x^2}{h} \).
This step sets the foundation for further simplification by expanding the terms.
Substitutions must be done carefully to ensure that the problem's logic is preserved. Each substitution brings us closer to a solution, simplifying the expression and making the subsequent steps much easier.

Expansion and Simplification

After substitution, expansion is necessary to break down expressions within parentheses. For our problem:

• Expand \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \)
• This leads to the expression \( \frac{(x^2 + 2xh + h^2) - x^2}{h} \)

The next part of the process is simplifying:

• Combine like terms by canceling out \( x^2 \)

This simplifies to \((2xh + h^2)/h \).
Expansion reveals the full content of expressions, allowing terms to be combined or canceled. It is fundamental in algebraic manipulation because it exposes the inner structure of expressions, revealing opportunities for further simplification.

Factoring

Factoring involves breaking down an expression into a product of simpler expressions. It's a technique used to simplify expressions and solve equations.

• In our problem, notice the common factor \( h \) in the numerator\((2xh + h^2)\)
• Factoring gives \( h(2x + h) \)

Once factored, you can cancel out \( h \) in the numerator and the denominator:

• This gives \( 2x + h \)

Factoring is crucial as it reduces expressions into simpler components. This simplification enables us to cancel common terms, solving equations more easily and deriving results that are manageable and understandable.