Problem 78 - A Graphical Approach to Precalculus with Limits

Question

Determine the difference quotient \(\frac{f(x+h)-f(x)}{h}\) (where \(h \neq 0\) j for each function \(f\). Simplify completely. $$f(x)=\frac{1}{2} x^{2}+4 x$$

Step-by-Step Solution

Answer

The simplified difference quotient is \( x + \frac{1}{2}h + 4 \).

Step 1: Substitute the function into the difference quotient

Given the function \( f(x) = \frac{1}{2} x^2 + 4x \), the difference quotient is \( \frac{f(x+h)-f(x)}{h} \). First, substitute \( f(x+h) \) and \( f(x) \) into the difference quotient formula:\[f(x+h) = \frac{1}{2} (x+h)^2 + 4(x+h)\]So the difference quotient becomes:\[\frac{\left(\frac{1}{2} (x+h)^2 + 4(x+h)\right) - \left(\frac{1}{2} x^2 + 4x\right)}{h}\]

Step 2: Expand and simplify \( f(x+h) \)

Calculate \( f(x+h) \) by expanding the terms.\[(x+h)^2 = x^2 + 2xh + h^2\]Substituting back:\[\frac{1}{2} (x+h)^2 + 4(x+h) = \frac{1}{2}(x^2 + 2xh + h^2) + 4x + 4h\]\[= \frac{1}{2}x^2 + xh + \frac{1}{2}h^2 + 4x + 4h\]

Step 3: Substitute back and simplify the difference quotient

Substitute the expanded form of \( f(x+h) \) back into the difference quotient:\[\frac{\left(\frac{1}{2}x^2 + xh + \frac{1}{2}h^2 + 4x + 4h\right) - \left(\frac{1}{2} x^2 + 4x\right)}{h}\]Simplify the expression:\[= \frac{\frac{1}{2}x^2 + xh + \frac{1}{2}h^2 + 4x + 4h - \frac{1}{2}x^2 - 4x}{h}\]Cancel out the \( \frac{1}{2}x^2 \) and \( 4x \):\[= \frac{xh + \frac{1}{2}h^2 + 4h}{h}\]

Step 4: Factor and cancel \( h \)

Notice that \( h \) is common in the numerator terms:\[= \frac{h(x + \frac{1}{2}h + 4)}{h}\]Since \( h eq 0 \), divide out \( h \):\[x + \frac{1}{2}h + 4\]

Step 5: Conclusion of the simplified difference quotient

The difference quotient simplifies to:\[x + \frac{1}{2}h + 4\]This is the complete simplification of the given difference quotient for the function \( f(x) = \frac{1}{2} x^2 + 4x \).

Key Concepts

Polynomial Functions

Polynomial functions involve sums of powers of variables with constant coefficients. These functions are among the more elementary components in algebra, making them an important building block in understanding higher mathematical concepts. In this specific problem, our polynomial is a quadratic function:

• It has a term with the power of 2, which is the quadratic term \( \frac{1}{2} x^2 \), where \( \frac{1}{2} \) is the coefficient.
• It also includes a linear term, \( 4x \), which represents a line with slope 4 in the plane.

Each term adds a different dimension to the behavior of the function when it is graphed. The quadratic term affects the curvature, while the linear part shifts it diagonally in the coordinate plane. Understanding these components helps when performing operations, such as calculating the difference quotient, as you can predict how these transformations will affect the result.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions to make them easier to understand and work with. In our solution, algebraic manipulation is crucial, especially when dealing with the difference quotient. Here's what happens:

• First, substitute \( x+h \) into the function, replacing original variable instances to find \( f(x+h) \). This involves expanding, as shown with \( (x+h)^2 = x^2 + 2xh + h^2 \).
• Next, simplify the expanded expressions. Each term needs to be carefully broken down and grouped to simplify them into a manageable form.
• Major manipulation occurs when combining like terms in the numerator to eventually cancel out terms, such as eliminating \( \frac{1}{2}x^2 \) and \( 4x \) from the numerator.

These steps demonstrate how algebraic manipulation transforms complex expressions into simpler forms, making it possible to extract meaningful information such as the simplified form of the difference quotient.

Limit Process

The concept of the limit process becomes crucial in understanding the behavior of functions, particularly when approximating instantaneous rates of change. In calculus, this is the basis for derivatives. When analyzing the difference quotient, the limit process is somewhat hinted at even if it isn't explicitly calculated in this exercise. Here's why it matters:

• As \( h ightarrow 0 \), the difference quotient \( \frac{f(x+h)-f(x)}{h} \) provides a better approximation of the derivative or the slope of the tangent line to the function at any given point \( x \).
• This process forms the foundation for differential calculus, where instantaneous changes and slopes are important.
• In this problem, even though the limit process isn't explicitly solved, understanding the principles helps in comprehending why the difference quotient simplifies as it does.

By practicing these steps often, you get more comfortable with the abstract concept of limits and how they work under the hood, preparing you for more complex calculus applications.