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)=x+\sin (2 x), \quad x_{0}=\pi / 2 \end{equation}

Step by step solution

01

Plot the Function

First, we need to plot the function \( y = f(x) = x + \sin(2x) \) over the interval \((x_0-1/2) \leq x \leq (x_0+3)\), where \( x_0 = \frac{\pi}{2} \).Calculate the interval:\[ x_0 - \frac{1}{2} = \frac{\pi}{2} - \frac{1}{2} \approx 1.07 \]\[ x_0 + 3 = \frac{\pi}{2} + 3 \approx 4.57 \]Now, the interval becomes \(1.07 \leq x \leq 4.57\). Use a computer algebra system (CAS) to plot the function over this interval.

02

Difference Quotient Function

The difference quotient is defined as \( q(h) = \frac{f(x_0 + h) - f(x_0)}{h} \).For \( f(x) = x + \sin(2x) \) and \( x_0 = \frac{\pi}{2} \), substitute into the function:\[ q(h) = \frac{f\left(\frac{\pi}{2} + h\right) - f\left(\frac{\pi}{2}\right)}{h} \]Evaluating the expressions:\[ f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \sin(\pi) = \frac{\pi}{2} \] (because \( \sin(\pi) = 0 \))\[ f\left(\frac{\pi}{2} + h\right) = \left(\frac{\pi}{2} + h\right) + \sin\left(2 \left(\frac{\pi}{2} + h\right)\right) \]\[ = \frac{\pi}{2} + h + \sin(\pi + 2h) = \frac{\pi}{2} + h - \sin(2h) \] (using the identity \( \sin(\pi + \theta) = -\sin(\theta) \))So, \[ q(h) = \frac{\left(\frac{\pi}{2} + h - \sin(2h)\right) - \frac{\pi}{2}}{h} = \frac{h - \sin(2h)}{h} = 1 - \frac{\sin(2h)}{h} \]

03

Find the Limit

We find the limit of \( q(h) \) as \( h \to 0 \):Calculate \( \lim_{h \to 0} q(h) = \lim_{h \to 0} \left(1 - \frac{\sin(2h)}{h}\right) \).Use L'Hôpital's Rule for \( \frac{\sin(2h)}{h} \):\[ \lim_{h \to 0} \frac{\sin(2h)}{h} = \lim_{h \to 0} \frac{2\cos(2h)}{1} = 2 \times \cos(0) = 2 \times 1 = 2 \]So, \[ \lim_{h \to 0} q(h) = 1 - 2 = -1 \]. The limit is \(-1\), the slope of the tangent line at \( x = x_0 \).

04

Define and Graph Secant Lines

Define the secant lines using the formula:\[ y = f(x_0) + q(h) \cdot (x - x_0) \]For various \( h \) values:- \( h = 3 \): \[ q(3) = 1 - \frac{\sin(6)}{3} \]- \( h = 2 \): \[ q(2) = 1 - \frac{\sin(4)}{2} \]- \( h = 1 \): \[ q(1) = 1 - \frac{\sin(2)}{1} \]Calculate each and plot these secant lines along with the original function \( f(x) \) and the tangent line (from Step 3) on the interval \(1.07 \leq x \leq 4.57\) using a CAS.

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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 foundational concept in calculus that helps measure how a function changes as its input changes. Essentially, it gives us the average rate of change of the function over an interval. Mathematically, it is expressed as follows:
\[ q(h) = \frac{f(x_0 + h) - f(x_0)}{h} \]This formula measures the change in the function value from \( x_0 \) to \( x_0 + h \). The difference quotient depends on the "step size" \( h \); smaller values of \( h \) provide greater insight into the behavior of the function near \( x_0 \).

When \( h \) approaches zero, the difference quotient approaches the derivative of the function at \( x_0 \), indicating the instantaneous rate of change or the slope of the tangent line at this point. This concept is crucial when analyzing function behavior and forming the basis for mathematical tools like derivatives, enabling deeper understanding of calculus.

Tangent Lines

A tangent line is a straight line that 'touches' a curve at a single point and has the same slope as the curve at that point. In calculus, tangent lines are used to approximate the value of a function near that point.

To find the equation of a tangent line at a point \( x_0 \), one must calculate the slope using the limit of the difference quotient as \( h \) approaches zero. For the function \( f(x) = x + \sin(2x) \) at \( x_0 = \frac{\pi}{2} \), the slope is found by evaluating:
\[ \lim_{h \to 0} q(h) = \lim_{h \to 0} \left(1 - \frac{\sin(2h)}{h}\right) = -1 \]This limit tells us that the slope of the tangent line at \( x_0 \) is \(-1\).

The equation of the tangent line is then determined using the point-slope form of a line:
\[ y = f(x_0) + m(x - x_0) \]where \( m \) is the slope, giving us an essential tool for linear approximations in calculus.

Secant Lines

Secant lines are lines that intersect a curve at two or more points, unlike tangent lines which meet the curve at just one point. These lines are crucial in understanding how functions behave over an interval.

The equation of a secant line through two points on a curve is given by the change of the function between those two points. For the function \( f(x) = x + \sin(2x) \), especially when evaluated at \( x_0 = \frac{\pi}{2} \), secant lines can be constructed for various values of \( h \) using:
\[ y = f(x_0) + q(h) \cdot (x - x_0) \]Calculating for specific values like \( h = 1, 2, 3 \), you get different slopes, creating various secant lines corresponding to these steps.

• For \( h = 1 \), the slope is calculated as \( 1 - \frac{\sin(2)}{1} \).
• For \( h = 2 \), it changes to \( 1 - \frac{\sin(4)}{2} \).
• For \( h = 3 \), it further evolves to \( 1 - \frac{\sin(6)}{3} \).

By graphing these secant lines alongside the tangent line and the function itself, students can visually appreciate the transition from average to instantaneous rates of change, enhancing their problem-solving skills in calculus.