Question

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

Short answer

Based on the given information, we can make a conjecture that the value of \(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\) is 1. This is because, from the completed table, as the values of \(x\) approach 0 (from both left and right sides), the values of \(\frac{3^x - 1}{x}\) become very close to 1.

Step-by-step solution

Sketch the graph of \(y=3^x\) and draw the secant lines

To sketch the graph of \(y=3^x\), start by plotting points for different values of \(x\). The given point \(P(0,1)\) is already on the graph as \(3^0=1\). Now, for drawing the secant lines connecting the points \(P\) and \(Q\) for the values of \(x=-2,-1,1\) and 2, plot the points \(Q(-2, 3^{-2})\), \(Q(-1, 3^{-1})\), \(Q(1, 3^1)\), and \(Q(2, 3^2)\). Then, draw lines connecting point \(P\) with each of the four points \(Q\).

Find the slope of the line connecting \(P\) and \(Q\)

The slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) can be found using the formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Given the point \(P(0,1)\) and \(Q(x,3^x)\), the slope of the secant line is: $$ m = \frac{3^x - 1}{x - 0} = \frac{3^x - 1}{x} $$ For \(x \neq 0\), the slope of the line connecting \(P\) and \(Q\) is \(m = \frac{3^x - 1}{x}\).

Complete the table and make a conjecture about the limit value

To complete the table, we need to compute the values of \(\frac{3^x - 1}{x}\) for each given value of \(x\): $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline x & -0.1 & -0.01 & -0.001 & -0.0001 & 0.1 & 0.01 & 0.001 & 0.0001 \\ \hline \frac{3^x-1}{x} & 1.10 & 1.01 & 1.001 & 1.0001 & 1.09 & 1.009 & 1.0009 & 1.00009 \\ \hline\end{array}$$ As the values of \(x\) approach 0 (from both left and right sides), the values of \(\frac{3^x - 1}{x}\) become very close to 1. We can make a conjecture that the value of \(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\) is 1.

Key concepts

Exponential functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. The general form is expressed as \(y = a^x\), where \(a\) is a positive constant different from zero, and \(x\) is the exponent, which is usually a real number.
The exponential function discussed in the exercise is \(y = 3^x\). This function represents a rapid rate of change as \(x\) increases, producing a curve that rises steeply on a graph as \(y\) becomes large.

• For positive values of \(x\), \(y = 3^x\) grows exponentially as \(3^x\) generates larger numbers.
• For negative values of \(x\), \(3^x\) results in fractions, as the reciprocal of powers of three will be less than 1.
• The graph of an exponential function is always above the \(x\)-axis, showing it never becomes zero.

Understanding this exponential growth is key in various fields, such as finance, biology, and physics, where growth processes occur exponentially.

Limits

Limits in calculus are fundamental in understanding how functions behave as inputs approach certain particular values, especially those of infinity or zero.
In the exercise provided, students are tasked with finding the limit \(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\). This expression represents a sequence of slopes for secant lines connecting two points as \(x\) nears zero.

• The table values suggest that as \(x\) goes towards zero, \(\frac{3^x - 1}{x}\) converges towards 1.
• This limit portrays the behavior of the function \(3^x\) as the gap between two points becomes infinitesimally small.

Using limits, students can understand both instantaneous rates of change (derivatives) and cumulative quantities (integrals). Evaluating limits such as these provide a foundation for developing a deeper understanding of calculus and its applications.

Tangent lines

A tangent line is a line that contacts a curve at a single point without crossing it over a small range around that point. It illustrates the concept of instantaneous rate of change.
In the exercise, students compute the slope of secant lines and find the tangent line for \(y = 3^x\) at the point \(P(0, 1)\) by calculating \(\lim_{x \rightarrow 0} \frac{3^x - 1}{x}\). These secant lines approach the tangent line as the separation between the point \(P\) and \(Q\) becomes nearly zero.

• The slope of the tangent line gives the derivative of the function at that point.
• Here, it's determined that the slope approaches 1 as \(x\) nears zero.
• The tangent line is vital in calculus as it represents the linear approximation of the function near a given point.

Understanding tangent lines is crucial for interpreting rates of change and essential for solving many real-world problems involving motion, growth, and decay.