91Ó°ÊÓ

Use a CAS to perform the following steps for the functions: \begin{equation}\begin{array}{l}{\text { a. Plot } y=f(x) \text { over the interval }\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)} \\ {\text { b. Holding } x_{0} \text { fixed, the difference quotient }}\end{array}\end{equation}\begin{equation} q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}\end{equation} at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. \begin{equation} \begin{array}{l}{\text { c. Find the limit of } q \text { as } h \rightarrow 0 \text { . }} \\ {\text { d. Define the secant lines } y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right) \text { for } h=3,2} \\ {\text { and } 1 . \text { Graph them together with } f \text { and the tangent line over }} \\ {\text { the interval in part (a). }}\end{array}\end{equation} \begin{equation} f(x)=\cos x+4 \sin (2 x), \quad x_{0}=\pi \end{equation}

Step by step solution

01

Define the Function

The function provided is \( f(x) = \cos x + 4\sin(2x) \). We are examining this function over a specific interval centered around \( x_0 = \pi \).

02

Determine the Interval for Plotting

We need to plot \( y = f(x) \) over the interval \( \left(x_0 - \frac{1}{2}\right) \leq x \leq \left(x_0 + 3\right) \). For \( x_0 = \pi \), this interval becomes \( \left(\pi - \frac{1}{2}\right) \leq x \leq (\pi + 3) \). Evaluate numerically: \( 2.6416 \leq x \leq 6.1416 \).

03

Difference Quotient Function

Define the difference quotient \( q(h) = \frac{f(x_0 + h) - f(x_0)}{h} \). Substitute \( f(x) = \cos x + 4\sin(2x) \) and \( x_0 = \pi \) to get: \[ q(h) = \frac{\cos(\pi + h) + 4\sin(2(\pi + h)) - (\cos \pi + 4\sin(2\pi))}{h} \].

04

Simplify the Difference Quotient

Substitute the known trigonometric values \( \cos \pi = -1 \) and \( \sin 2\pi = 0 \) to simplify: \[ q(h) = \frac{(\cos(\pi + h) + 4\sin(2\pi + 2h) + 1)}{h} \].

05

Limit of the Difference Quotient

Find the limit of \( q(h) \) as \( h \rightarrow 0 \). Calculating this limit gives the derivative of \( f(x) \) at \( x = \pi \). After evaluating limits: \[ \lim_{h \to 0} q(h) = -1 \].

06

Define Secant Lines

Calculate secant lines using \( h = 3, 2, 1 \). The general equation is: \[ y = f(x_0) + q(h) \cdot (x-x_0) \]. Evaluate \( f(\pi) = \cos \pi + 4\sin 2\pi = -1 \). Substitute into the equation for each \( h \).

07

Graph Functions

In a CAS, plot \( y = f(x) \) and the secant lines for \( h = 3, 2, 1 \), as well as the tangent line \( y = f(\pi) - (x - \pi) \) over the interval \( 2.6416 \leq x \leq 6.1416 \). Show how the secant lines approach the tangent as \( h \rightarrow 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference Quotient

The difference quotient is a fundamental concept in calculus that helps us understand how functions change. It's a mathematical expression that approximates the slope of the function as it changes over a small distance. The formula for the difference quotient is given as: \( q(h) = \frac{f(x_0 + h) - f(x_0)}{h} \) In simpler terms, the difference quotient measures the average rate of change of the function \( f(x) \) between two points \( x_0 \) and \( x_0 + h \). You can think of it as a tiny "slope" of a line through these points. As we make \( h \) closer to zero, this average rate of change becomes more precise. Using the function from our problem, \( f(x) = \cos x + 4\sin(2x) \), we substitute the \( x_0 = \pi \) and find the expression for \( q(h) \). Substitutions are made using known trigonometric values to simplify the expression, aiding in understanding and solving further steps.

Secant Line

The secant line connects two points on the curve of a function. It helps you visualize and understand average changes in the function over an interval. The formula for a secant line is derived from the difference quotient. The equation for a secant line through points\( (x_0, f(x_0)) \) and \( (x_0 + h, f(x_0+h)) \) can be given as: \[ y = f(x_0) + q(h) \cdot (x - x_0) \] In our case, secant lines are calculated for specific step sizes \( h \) like 3, 2, and 1. We calculate each line by substituting these values into the secant line formula to see how close each line comes to the tangent as \( h \) reduces. These secant lines are graphed to visualize their approach towards the tangent line.

Tangent Line

The tangent line is a concept used to describe the slope at an exact point on a function's curve. Unlike the secant line, which averages the slope over an interval, the tangent line gives the slope at a specific point. In calculus, the derivative of a function at a point gives us the slope of the tangent line there. In this exercise, when \( h \rightarrow 0 \), the secant line becomes indistinguishable from the tangent line, showing that the difference quotient turns into the derivative. For the function \( f(x) = \cos x + 4\sin(2x) \) at \( x_0 = \pi \), we calculated that the limit of the difference quotient or the derivative is \(-1\). Therefore, the tangent line at \( x = \pi \) is: \[ y = f(x_0) - (x - x_0) \] Essentially, it confirms the established idea that the tangent line represents instantaneous rates of change.

Trigonometric Functions

Trigonometric functions like sine and cosine are periodic and oscillate in predictable patterns. In calculus, they have derivatives that help us analyze and predict their behavior on a graph. In our problem, the function provided \( f(x) = \cos x + 4\sin(2x) \) is a combination of the cosine and sine functions. Cosine generally represents horizontal waves, while sine represents vertical waves. When combined, they create complex wave-like patterns.

• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \sin x \) is \( \cos x \).

Understanding these derivatives is crucial as they form the basis for much of calculus involving trigonometric functions. They help us to calculate rates of change and slopes across their oscillations, as evidenced by our calculations of derivatives in the exercise.