Question

In Exercises 4 and \(42,\) complete the following for the function. (a) Compute the difference quotient \(\frac{f(1+h)-f(1)}{h}\) (b) Use graphs and tables to estimate the limit of the difference quotient in part (a) as \(h \rightarrow 0\) . (c) Compare your estimate in part (b) with the given number. (d) Writing to Learn Based on your computations, do you think the graph of \(f\) has a tangent at \(x=1 ?\) If so, estimate its slope. If not, explain why not. \(f(x)=2^{x}, \quad \ln 4\)

Short answer

The difference quotient is \(\frac{2^{1+h}-2}{h}\). As \(h\) approaches \(0\), the limit of the difference quotient is approximately \(\ln 4\), which also equals to the given number. This indicates that the graph of \(f(x)\) has a tangent at \(x=1\) and its slope is \(\ln 4\).

Step-by-step solution

Compute the difference quotient

Substitute \(x = 1\) and \(x = 1 + h\) into \(f(x) = 2^x\) to find \(f(1+h)\) and \(f(1)\). Then, compute the difference quotient \(\frac{f(1+h)-f(1)}{h}\)

Estimate the limit of the difference quotient

Create a table with decreasing values of \(h\) and corresponding difference quotient values. As \(h\) approaches \(0\), observe and estimate the limit of the difference quotient.

Compare estimates with the given number

Compare the limit estimated in step 2 with the given number, \(\ln 4\). They should be approximately equal, as \(\ln 4\) is the derivative of \(2^x\) at \(x=1\)

Determine if the graph of \(f\) has a tangent at \(x=1\) and estimate its slope

The graph of \(f\) does have a tangent at \(x=1\). The slope of the tangent line at \(x=1\) is the limit of the difference quotient as \(h \rightarrow 0\), which is estimated in step 2.

Key concepts

Limit of Difference Quotient

The limit of the difference quotient is an essential concept in calculus, particularly for understanding the derivative's foundation. When we compute the difference quotient \(\frac{f(x+h)-f(x)}{h}\), we're measuring the average rate of change of the function \(f\) over an interval \(h\). However, what we're really interested in is the instantaneous rate of change, which is what a derivative represents.

The difference quotient becomes an approximation of the derivative as \(h\) gets closer to zero. To find the derivative at a point \(x\), we take the limit of the difference quotient as \(h\) approaches zero (\(h \rightarrow 0\)). In our exercise, the difference quotient \(\frac{f(1+h)-f(1)}{h}\) is evaluated as \(h \rightarrow 0\) to estimate the derivative of \(f(x) = 2^x\) at \(x = 1\). This limit, when computed, should match the known derivative at that point, \(\ln(4)\).

Approaching Zero

To elaborate, as \(h\) decreases in size, we're essentially squeezing the interval over which we measure the average rate of change until it provides the exact slope of the tangent line at that point. Tables and graphs are helpful tools for visualizing how the difference quotient values converge as \(h\) shrinks. This process helps students understand not just the numerical process, but also the underlying concept of the limit that defines the derivative.

Tangent Line Slope Estimation

In calculus, the tangent line to a curve at a particular point is the line that 'just touches' the curve at that point. The slope of this tangent line is particularly important as it represents the derivative of the function at that point. Estimating the slope of a tangent line is closely related to understanding the limit of the difference quotient.

Using the given function \(f(x) = 2^x\), our goal is to determine whether \(f\) has a tangent at \(x=1\) and to estimate this line's slope. Utilizing the procedures outlined in our solution, which includes computing the difference quotient and observing its behavior as \(h\) approaches zero, provides an estimate for this slope. The estimation is based on the recognition that as \(h \rightarrow 0\), the slope of the secant lines (lines connecting two points on a curve) approaches the slope of the tangent line.

Visualizing with Graphs

Graphs can make this estimation more intuitive by showing the secant lines becoming steeper and closer to the tangent line as \(h\) decreases. This graphical approach aids students in visualizing how secant lines converge towards the tangent as their shared point gets infinitesimally close, thus growing their understanding of the concept.

Computing Derivatives

Computing derivatives is a fundamental operation in calculus, allowing us to determine the instantaneous rate of change of a function at any given point. The derivative is defined as the limit of the difference quotient as \(h\) approaches zero. In practice, this involves computing limits, applying differentiation rules, or using the definition of derivative with the difference quotient.

The given exercise requires us to compute the derivative of \(f(x) = 2^x\) using the difference quotient. In general, to compute derivatives of more complex functions, one might need to apply various rules such as the product rule, quotient rule, chain rule, or power rule. These rules simplify the process of differentiation and make it manageable to find derivatives of more complicated expressions.

Derivative Shortcuts

While the difference quotient gives the derivative from first principles, with practice, students learn shortcuts to speed up the process. For example, they will learn that the derivative of \(2^x\) is \(2^x \ln(2)\), which simplifies the process significantly compared to working through the limit of the difference quotient every time.

Derivative at a Point

The concept of the derivative at a specific point is integral to the study of calculus. This tells us the exact rate at which a function is changing at one particular point. In the exercise, we're interested in finding the derivative of the function \(f(x) = 2^x\) when \(x=1\).

To calculate the derivative at \(x=1\), we use the limit of the difference quotient method as outlined in the step-by-step solution. The numerical approximation we obtain through the calculations should agree with the established derivative value for \(2^x\) at \(x=1\), which is \(\ln(4)\).

Understanding the Derivative

Understanding how to find the derivative at a point helps to conceptualize what it means for a function to have a 'speed' at a certain instant. It enables students to grasp the importance of derivatives in real-world applications, such as physics, where it can represent velocities or rates of change in various contexts. Mastering this concept equips students to delve into further applications of calculus, including optimization and motion analysis.