Question

Slope of a tangent line a. Sketch a graph of \(y=2^{x}\) and carefully draw three secant lines connecting the points \(P(0,1)\) and \(Q\left(x, 2^{x}\right),\) for \(x=-3,-2,\) and -1 b. Find the slope of the line that passes through \(P(0,1)\) and \(Q\left(x, 2^{x}\right),\) for \(x \neq 0\). c. Complete the table and make a conjecture about the value of \(\lim _{x \rightarrow 0^{-}} \frac{2^{x}-1}{x}\). $$\begin{array}{|l|l|l|l|l|l|l|} \hline x & -1 & -0.1 & -0.01 & -0.001 & -0.0001 & -0.00001 \\ \hline \frac{2^{x}-1}{x} & & & & & & \\ \hline \end{array}$$

Short answer

Answer: As x approaches 0 from the negative side, the conjectured value of the limit for the expression (2^x - 1)/x is approximately 0.6923.

Step-by-step solution

(Part a: Sketch the graph)

To sketch the graph of \(y=2^x\), plot a few points on the graph and connect them with a smooth curve. Carefully draw three secant lines connecting the points \(P(0,1)\) and \(Q(x, 2^x)\) for \(x=-3, -2\), and -1.

(Part b: Find the slope of the line that passes through P and Q)

To find the slope of the line that passes through \(P(0,1)\) and \(Q(x, 2^x)\), use the slope formula: \(m = \frac{y_2 - y_1}{x_2 - x_1}\) Here, \(x_1 = 0\), \(y_1 = 1\), \(x_2=x\), and \(y_2=2^x\). Plug in the values and get: \(m = \frac{2^x - 1}{x - 0} = \frac{2^x - 1}{x}\) This is the slope of the line that passes through \(P\) and \(Q\) for \(x\neq 0\).

(Part c: Complete the table and conjecture about the limit)

To complete the table, evaluate the expression \(\frac{2^x - 1}{x}\) for each given value of \(x\). After completing the table, observe the values and make a conjecture about the value of the limit as \(x \rightarrow 0^{-}\): $$\begin{array}{|l|l|l|l|l|l|l|} \hline x & -1 & -0.1 & -0.01 & -0.001 & -0.0001 & -0.00001 \\ \hline \frac{2^{x}-1}{x} & \frac{1}{2} & \approx 0.69 & \approx 0.6923 & \approx 0.69231& \approx 0.692307 & \approx 0.6923077 \\ \hline \end{array}$$ We can observe that as \(x\) approaches \(0^{-}\), the value of \(\frac{2^{x}-1}{x}\) seems to approach \(\log_{e}{2}\), which is approximately \(0.6923\). Thus, we can make a conjecture that: \(\lim _{x \rightarrow 0^{-}} \frac{2^{x}-1}{x} \approx 0.6923\)

Key concepts

Secant Lines

In the study of calculus, secant lines are essential tools for understanding the behavior of curves at a particular point. Imagine a curve on a graph, and pick two points on that curve. The line that connects these two points is known as a secant line. These lines are especially useful when you're interested in finding the average rate of change of the function between two points.

In the context of our exercise, three secant lines are drawn from a common point, P(0,1), to three other points on the curve of an exponential function, each secant line representing an average rate of change over a different interval. As the two points get closer together, the secant line becomes a better approximation for the tangent line at point P, which represents the instantaneous rate of change or the slope of the curve at that point.

Slope Formula

The slope of a line is a measure of its steepness and direction. The slope is found using the slope formula given by:
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
This formula represents the change in the y-values, known as the rise, over the change in the x-values, known as the run. When working with secant lines, the slope formula gives us the average rate of change between two points. Calculating the slope for secant lines on our graph, as in the textbook exercise, sets the stage for understanding the slope of the tangent line, a concept fundamental to differential calculus.

Limit of a Function

The concept of a limit is central to calculus and analysis, dealing with the behavior of a function as the input approaches a particular point. Limits help in studying the change of a function at a specific point, even when the function itself isn't defined at that point.

In our exercise, we look at the limit of a function as the x-value approaches 0 from the left (denoted as \( x \rightarrow 0^- \)). By filling out the table and calculating the value of \( \frac{2^x - 1}{x} \) for values of x closer and closer to zero, we are observing the behavior of the secant slopes as they approach the slope of the tangent line at P. The conjecture about the limit is informed by these observations, signaling what the slope would be at an infinitesimally small interval around P.

Exponential Functions

Exponential functions are characterized by their variable exponent. For instance, in the function \( y=2^x \), 2 is the base and x is the exponent. These functions are widely applicable across various scientific fields, representing rapid growth or decay processes.

In the exercise, we consider the exponential function \( y=2^x \) to study the behavior of the rate of change as we look at different intervals. As depicted by the graph and secant lines, exponential functions increase or decrease at rates proportional to their current value, which is why understanding their slopes at specific points provides insight into the functions' behavior at those points. The slope of the tangent line, in this case, provides information about the instantaneous growth rate of the function at the point P(0,1).