Problem 39 - Calculus

Question

Draw \(1 / x\) and \(1 /(x+h)\) and \(\Delta f / h\) -either by hand with \(h=\frac{1}{2}\) or by computer to show \(h \rightarrow 0\).

Step-by-Step Solution

Answer

The derivative as \( h \rightarrow 0 \) is \( \frac{-1}{x^2} \), demonstrated by the limit of the difference quotient.

Step 1: Understanding the Functions

Let's first understand the functions we are dealing with. The functions are \( f(x) = \frac{1}{x} \) and \( f(x + h) = \frac{1}{x+h} \). We plot these two functions with a specific value of \( h \), which is given as \( h = \frac{1}{2} \).

Step 2: Define the Difference Quotient

The given expression \( \Delta f / h \) represents the difference quotient, which is defined as \[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \].This expression will help approximate the derivative of the function \( f(x) = \frac{1}{x} \) as \( h \rightarrow 0 \).

Step 3: Simplify the Difference Quotient

To simplify the expression, we need to find a common denominator for the terms in the numerator. \[ \frac{1}{x+h} - \frac{1}{x} = \frac{x - (x+h)}{x(x+h)} = \frac{-h}{x(x+h)} \].Thus, the difference quotient becomes\[ \frac{-h}{h \cdot x(x+h)} = \frac{-1}{x(x+h)} \].

Step 4: Plotting the Functions

Using graph paper or computer software, plot \( f(x) = \frac{1}{x} \) and \( f(x+h) = \frac{1}{x+h} \) for \( h = \frac{1}{2} \). Notice how the graph of \( f(x+h) \) is slightly shifted from \( f(x) \). These shifts get smaller as \( h \rightarrow 0 \).

Step 5: Visualize the Difference Quotient

Compute and plot \( \Delta f / h \) from the simplified expression \( \frac{-1}{x(x+h)} \). Show how this quotient represents the slope of the tangent line to the curve \( y = \frac{1}{x} \) as \( h \rightarrow 0 \). As \( h \) approaches zero, \( \Delta f / h \) approximates the derivative \( \frac{d}{dx}\frac{1}{x} \).

Step 6: Conclusion on the Derivative

As \( h \rightarrow 0 \), the difference quotient \( \Delta f / h = \frac{-1}{x(x+h)} \) approaches \( \frac{-1}{x^2} \), which is the derivative of \( \frac{1}{x} \). This demonstrates the concept of derivatives via the limit of the difference quotient.

Key Concepts

Difference Quotient

The difference quotient is a fundamental concept in calculus that helps us understand how functions change. It's given by the formula:

• \(\frac{f(x+h) - f(x)}{h}\)

If you have a function, say \( f(x) = \frac{1}{x} \), the difference quotient for this function would become:

• \(\frac{\frac{1}{x+h} - \frac{1}{x}}{h}\)

This is essential for finding instantaneous rates of change, aka derivatives. By plugging the terms into the formula and simplifying, you establish how changes in \(x\) affect \(f(x)\). The simpler form for \( f(x) = \frac{1}{x} \) with difference quotient computation is:

• \(\frac{-1}{x(x+h)}\).

Limit Definition of the Derivative

The limit definition of the derivative uses the difference quotient to find an exact slope of a tangent line to a function at a specific point. It's expressed as:

• \( \lim_{{h \to 0}} \frac{f(x+h) - f(x)}{h} \)

This limit describes how the difference quotient behaves as \( h \) approaches zero. Essentially, you are evaluating what happens when two points on a curve are brought infinitely close together.In the context of \( f(x) = \frac{1}{x} \), the limit becomes:

• \( \lim_{{h \to 0}} \frac{-1}{x(x+h)} \),

which simplifies to the derivative \( \frac{-1}{x^2} \).By using the limit definition, you grasp that derivative as a slope that tells how steep the curve is at a point. This slope is crucial for understanding change and predictions of function behavior. In calculus, this helps decrypt complex changes of functions in various scientific fields.

Function Transformation

Function transformation involves altering a given function to understand different behaviors. For the function \( f(x) = \frac{1}{x} \), replacing \( x \) with \( x + h \) results in:

• \( f(x+h) = \frac{1}{x+h} \)

This transformation shifts the graph horizontally. Evaluating the impact of each transformation shows how small shifts affect the curve.Specifically, plotting \( \frac{1}{x} \) and \( \frac{1}{x+h} \) helps identify how the graph shifts with changes in \( h \). Observing these shifts as \( h \to 0 \) helps reveal a fundamental behavior: the functions converge, and gaps narrow, a key insight when examining derivatives.Function transformations are powerful tools, allowing one to view how changes modify graphs. By mastering these, analyzing and graphing intricate functions becomes more intuitive, enabling a deeper understanding of changes and shifts in mathematical models.

Visualization in Calculus

Visualization is crucial in calculus, as it aids in understanding abstract concepts like differentiation. By graphing \( f(x) = \frac{1}{x} \) and its transformations such as \( f(x+h) = \frac{1}{x+h} \), one sees the effect as \( h \to 0 \).Visual aids include plotting the difference quotient \( \frac{-1}{x(x+h)} \) to see how it converges to the derivative \( \frac{-1}{x^2} \) as \( h \to 0 \). Graphically, this represents the slopes of tangent lines becoming more visible and precise.Visualizing these graphs makes understanding slopes and derivatives clearer. Techniques like using graph paper or software can help you see the mathematical concepts in action. The more you visualize, the easier it is to grasp why derivatives work mathematically, providing insight into real-function behaviors.

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Problem 40

Other exercises in this chapter

Problem 39

True or false, with a good reason: (a) The derivative of \(x^{2 n}\) is \(2 n x^{2 n-1}\). (b) By linearity the derivative of \(a(x) u(x)+b(x) v(x)\) is \(a(x)

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Problem 39

At a distance \(\Delta x\) from \(x=2,\) how far is the curve \(y=x^{3}\) above its tangent line?

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Problem 40

Name a product whose price elasticity is (a) high (b) low (c) negative (?)

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Problem 40

For \(y=e^{x},\) show on computer graphs that \(d y / d x=y\).

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