Problem 17 - Precalculus

Question

Find the limit of the difference quotient for each function \(f(x)\) given, to obtain a function \(f(x)\) that represents the instantaneous rate of change at \(x\) for each function. $$f(x)=\frac{1}{2} x+5$$

Step-by-Step Solution

Answer

The instantaneous rate of change is \( \frac{1}{2} \).

Step 1: Define the Difference Quotient

The difference quotient for a function \( f(x) \) is given by \( \frac{f(x+h) - f(x)}{h} \), where \( h \) is a small increment. In this case, the function is \( f(x) = \frac{1}{2}x + 5 \).

Step 2: Substitute and Simplify

Substitute \( f(x) \) into the difference quotient formula:\[\frac{f(x+h) - f(x)}{h} = \frac{\left(\frac{1}{2}(x+h) + 5\right) - \left(\frac{1}{2}x + 5\right)}{h}\]Simplify the expression:\[= \frac{\frac{1}{2}x + \frac{1}{2}h + 5 - \frac{1}{2}x - 5}{h}\]\[= \frac{\frac{1}{2}h}{h}\]

Step 3: Cancel Terms and Simplify Further

Notice in the numerator that the terms \( \frac{1}{2}x \) and \( 5 \) cancel out. Simplify further:\[= \frac{1}{2}\]Since \( h eq 0 \), the difference quotient reduces to a constant.

Step 4: Evaluate the Limit as h Approaches Zero

The last step in finding the derivative is to take the limit of the difference quotient as \( h \) approaches zero.\[\lim_{{h \to 0}} \frac{1}{2} = \frac{1}{2}\]

Step 5: State the Derivative

The derivative of \( f(x) = \frac{1}{2}x + 5 \), which represents the instantaneous rate of change, is the constant \( \frac{1}{2} \). This means that the rate of change does not depend on \( x \) and is constant for this linear function.

Key Concepts

Instantaneous Rate of Change

Imagine you're on a journey and you want to know how fast you're traveling at a specific moment. The "instantaneous rate of change" is precisely this snapshot of speed at one instant in time for functions. It tells us how quickly a function's value is changing at any given point.

• This is different from finding the average rate of change, which measures the speed over an interval. Instantaneous rate of change hones in on a single point.
• To find this rate, we use a method involving the "difference quotient," which compares changes in function's value over an infinitesimally small interval.

Understanding this concept is fundamental in calculus, as it opens doors to analyzing dynamic systems, motion, growth rates, and much more. Here, for the function \( f(x) = \frac{1}{2}x + 5 \), its instantaneous rate of change is a constant value indicating uniform speed or change.

Derivative

The derivative is a core concept in calculus, functioning as the tool we use to compute the instantaneous rate of change. It's what we get when we take the limit of the difference quotient as the interval approaches zero.

• Think of the derivative as the ultimate rate indicator, giving us insights into the behavior of functions precisely at each point.
• Mathematically, if you have a function \( f(x) \), its derivative \( f'(x) \) symbolizes the function's slope or rate of change at any given \( x \).

For the linear function \( f(x) = \frac{1}{2}x + 5 \), applying the derivative process reveals a rate of \( \frac{1}{2} \). This highlights how the function changes uniformly, matching our understanding of lines: they have constant slopes.

Linear Function

Linear functions are straightforward and foundational in mathematics. They describe relationships with a constant rate of change, represented graphically as straight lines. A linear function takes the form \( ax + b \), where \( a \) is the slope and \( b \) is the y-intercept.

• Slope \( a \) is crucial; it determines the tilt of the line. In our exercise, the function \( f(x) = \frac{1}{2}x + 5 \) has a slope of \( \frac{1}{2} \).
• This consistent slope indicates that the rate of change is steady across all values of \( x \), which is a hallmark of linear behavior.
• The y-intercept \( b \), in this case, 5, shows where the line crosses the y-axis.

Understanding linear functions equips you with the basics needed to tackle more complex functions in calculus. They provide a clear example of constant change and are the perfect backdrop for exploring varying rates in other types of functions.