Problem 106 - Algebra and Trigonometry Real Mathematics, Real People

Question

Consider the function \(f(x)=3 \sin (0.6 x-2)\) (a) Find the zero of \(f\) in the interval [0,6] (b) A quadratic approximation of \(f\) near \(x=4\) is \(g(x)=-0.45 x^{2}+5.52 x-13.70\) Use a graphing utility to graph \(f\) and \(g\) in the same viewing window. Describe the result. (c) Use the Quadratic Formula to approximate the zeros of \(g .\) Compare the zero in the interval [0,6] with the result of part (a).

Step-by-Step Solution

Answer

The zero of \(f(x)\) in the interval [0,6] is approximately \(x=5.24\). The quadratic function \(g(x)\) approximates the original function \(f(x)\) closely near \(x=4\). When we apply the quadratic formula to find the zeros of \(g(x)\), one root lies inside the interval [0,6] and is close to the zero from part (a), thus indicating a good approximation closer to the zero of \(f(x)\).

Step 1: Find the zero of f in the interval [0,6]

Setting \(f(x)=0\), we get: \(3\sin(0.6x-2) = 0\) Solving this for x in the interval [0,6], we get: \(\sin(0.6x-2)=0\) Let \(u=0.6x-2\), then \(\sin(u) = 0\) when \(u=n\pi\), where n is any integer. Considering our constraint, then \(0.6x-2 = n\pi \Rightarrow x = \frac{2+n\pi}{0.6}\) For \(n=1\) it is \(x = \frac{2+\pi}{0.6} \approx 5.24\) and lies inside the interval.

Step 2: Graph f and g and describe the result

Use a graphing utility to plot the given functions \(f(x)\) and \(g(x)\) in the same interval [0,6]. Observe how well the quadratic function \(g(x)\) approximates our original function \(f(x)\) near the point \(x=4.\)

Step 3: Apply Quadratic Formula to compute zeros of g and compare with 'a'

The quadratic function \(g(x)\) is given as \(g(x)=-0.45x^{2}+5.52x-13.70\). Solving for roots using the quadratic formula \(\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), we get the roots as \(x = \frac{-5.52 \pm \sqrt{5.52^{2} - 4*(-0.45)*(-13.70)}}{2*(-0.45)}\). Solving for \(x\), it will provide two roots, one inside and one outside the interval [0,6]. Compare the root lying inside the interval with the result from step 1. The two results should be close to one another.

Key Concepts

Quadratic Approximation

When we study functions, especially trigonometric ones like our given function of the form \(f(x) = 3 \sin (0.6 x-2)\), we are often interested in understanding their behavior near a specific point. Quadratic approximation is a mathematical method that allows us to use a simpler function to closely represent another, more complex function near a particular point - in this case, near \(x = 4\).

The quadratic function \(g(x)=-0.45 x^{2}+5.52 x-13.70\) is crafted to mirror the behavior of the original sine function around that specific value of \(x\). This approximation can greatly simplify our calculations and provide an intuitive understanding of the function's local behavior without having to deal with the complexity of trigonometric calculations.

Solving Trigonometric Equations

Trigonometric equations, such as \(3\sin(0.6x-2) = 0\), can usually be solved by using identities and properties of trigonometric functions. When we set the trigonometric function equal to zero, we're looking for the values of \(x\) where the function crosses the x-axis, also known as zeros or roots of the function.

To solve these equations, it is common to use the unit circle or trigonometric identities to find angles that make the function equal to zero. For the sine function, this occurs at multiples of \(\pi\) because sine is zero whenever its argument corresponds to the angle where the point on the unit circle crosses the x-axis.

Graphing Utility Analysis

Graphing utilities, like calculators and computer algebra systems, can be enormously helpful in visualizing functions and their approximations. By graphing our function \(f(x)\) and the quadratic approximation \(g(x)\) together, students can visually assess how well the approximation \(g(x)\) mimics \(f(x)\).

Besides offering visual insights, graphing utilities also allow for the evaluation of functions at numerous points to better understand the overall trends and behaviors, which may not be easily apparent from the equation alone. Through graphing, students can confirm the accuracy of the quadratic approximation in a specific interval, offering a powerful way to connect the algebraic manipulation to geometric interpretation.

Quadratic Formula Application

The quadratic formula \(\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) provides a systematic way to find the roots of any quadratic equation of the form \(ax^{2} + bx + c = 0\).

In our exercise, applying the quadratic formula to \(g(x)=-0.45x^{2}+5.52x-13.70\), allows students to approximate the zeros of \(g\) analytically. By solving the formula, students will find two solutions, one of which should fall within the interval [0,6] and closely match the zero of the original function \(f(x)\) found earlier. This exercise demonstrates a practical application of the quadratic formula and reinforces the connection between algebraic solutions and the graphical interpretations previously discussed.