Problem 40 - Precalculus with Calculus Previews

Question

Use a calculator to approximate the value \(m(b)=\lim _{h \rightarrow 0} \frac{b^{h}-1}{h}\) for \(b=1.5, b=2\), \(b=3\), and \(b=5\) by filling out the given table. $$ \begin{array}{|l|l|l|l|l|l|l|} \hline h \rightarrow 0 & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 & 0.000001 \\ \hline \frac{2^{h}-1}{h} & & & & & & \\ \hline \end{array} $$

Step-by-Step Solution

Answer

For \( b=2 \), \( m(2) \approx 0.72 \).

Step 1: Understand the Function m(b)

The function \( m(b) = \lim _{h \rightarrow 0} \frac{b^{h}-1}{h} \) represents the derivative of the exponential function \( b^h \) at \( h=0 \). This limit shows how the function \( b^h \) changes as \( h \) approaches zero. We're approximating this derivative for specific values of \( b \).

Step 2: Substitute Values for b

In sequence, we will substitute the values of \( b = 1.5, 2, 3, \) and \( 5 \) into the formula \( \frac{b^{h}-1}{h} \) to compute approximate values for small \( h \). Begin with \( b=2 \) and substitute into the expression.

Step 3: Calculate for h = 0.1

Substitute \( h = 0.1 \) into \( \frac{2^{h}-1}{h} \):\[ \frac{2^{0.1}-1}{0.1} = \frac{1.072-1}{0.1} = 0.72 \]

Step 4: Calculate for h = 0.01

Substitute \( h = 0.01 \):\[ \frac{2^{0.01}-1}{0.01} = \frac{1.0072-1}{0.01} = 0.72 \]

Step 5: Calculate for h = 0.001

Substitute \( h = 0.001 \):\[ \frac{2^{0.001}-1}{0.001} = \frac{1.00072-1}{0.001} = 0.72 \]

Step 6: Calculate for h = 0.0001

Substitute \( h = 0.0001 \):\[ \frac{2^{0.0001}-1}{0.0001} = \frac{1.000072-1}{0.0001} = 0.72 \]

Step 7: Calculate for h = 0.00001

Substitute \( h = 0.00001 \):\[ \frac{2^{0.00001}-1}{0.00001} = \frac{1.0000072-1}{0.00001} = 0.72 \]

Step 8: Calculate for h = 0.000001

Substitute \( h = 0.000001 \):\[ \frac{2^{0.000001}-1}{0.000001} = \frac{1.00000072-1}{0.000001} = 0.72 \]

Step 9: Complete the Table

Fill in the table with the calculated values for \( b=2 \) for each \( h \). The values are all approximately 0.72, showing the derivative approximation stabilizing.\[\begin{array}{|l|l|l|l|l|l|l|}\hline h \rightarrow 0 & 0.1 & 0.01 & 0.001 & 0.0001 & 0.00001 & 0.000001 \\hline \frac{2^{h}-1}{h} & 0.72 & 0.72 & 0.72 & 0.72 & 0.72 & 0.72 \\hline\end{array}\]

Key Concepts

Derivative approximation

When we talk about derivative approximation, we are essentially discussing the way we estimate the rate of change of a function. In calculus, derivatives represent this rate of change. For the function \( b^h \), we utilize the limit \( \lim _{h \rightarrow 0} \frac{b^{h}-1}{h} \) to approximate its derivative. The idea is to find how much the function \( b^h \) changes as the variable \( h \) becomes very, very small.For instance, consider the derivative at \( h = 0 \) which is where the rate of change is most accurately approximated. By plugging smaller and smaller values of \( h \) into the expression \( \frac{b^{h}-1}{h} \), we get closer to the actual derivative value. This approach is extremely useful for understanding how functions behave around specific points, especially when solving problems that involve exponential growth or decay, like in our exercise with different base values \( b \).

Exponential functions

Exponential functions are mathematical functions of the form \( b^x \), where \( b \) (the base) is a constant and \( x \) is a variable. These types of functions are characterized by their consistent growth rate when \( b \) is greater than one, making them incredibly useful in fields like finance, physics, and population ecology. The most familiar example is the natural exponential function \( e^x \), where \( e \) (approximately 2.71828) is an important mathematical constant.In our problem, we're analyzing exponential functions with different bases: \( 1.5, 2, 3, \) and \( 5 \). Each changes at a distinct rate as \( h \) approaches zero. Understanding these rates through derivative approximation helps us predict how quickly the function is increasing or decreasing at particular points. It’s the principle behind many real-world applications such as compound interest and radioactive decay.

• Base greater than 1: Function grows as \( x \) increases.
• Base equal to 1: Function remains constant.
• Base between 0 and 1: Function decreases as \( x \) increases.

Calculus limits

The concept of calculus limits is foundational for understanding calculus as a field. Limits help determine how a function behaves as it approaches a certain value or point. When we write \( \lim _{h \rightarrow 0} \frac{b^{h}-1}{h} \), we're looking at what happens to the fraction as \( h \) gets closer and closer to 0.Limits serve as the core idea that makes defining derivatives possible. Without limits, finding the instantaneous rate of change, or the slope of the tangent line at a single point, would be unfeasible. Applying limits to exponential functions, as in the exercise, allows us to comprehend not just simple growth patterns but detailed rates at which these patterns happen around specific points.
In our calculation for \( b = 2 \), by substituting even smaller numbers for \( h \), you might notice that the values of \( \frac{2^{h}-1}{h} \) stabilize around 0.72, illustrating the process of approaching the limit. This consistency indicates that our approximation is becoming increasingly precise as \( h \) nears zero, leading us to a clearer understanding of the derivative at that point.