Problem 28 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\cos x+\sqrt{3} \sin x, \quad 0 \leq x \leq 2 \pi$$

Step-by-Step Solution

Answer

Local and absolute extrema: max at \(x = \frac{\pi}{3}\), min at \(x = \frac{4\pi}{3}\). Inflection points at \(x = \frac{5\pi}{6}\) and \(x = \frac{11\pi}{6}\).

Step 1: Find the First Derivative

The first step to finding extreme points is to compute the first derivative of the function. The original function is \[ y = \cos{x} + \sqrt{3} \sin{x} \] The derivative is obtained by differentiating each term: \[ y' = -\sin{x} + \sqrt{3} \cos{x} \]

Step 2: Determine Critical Points

Critical points occur where the first derivative is zero or undefined. Set the first derivative equal to zero: \[ -\sin{x} + \sqrt{3} \cos{x} = 0 \] Simplifying gives: \[ \tan{x} = \sqrt{3} \] \( x_0 = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}, \frac{4\pi}{3} \) within the interval \( 0 \leq x \leq 2 \pi \).

Step 3: Determine Local and Absolute Extrema

Evaluate the original function at the critical points and endpoints to determine minima or maxima:- At endpoints: - \( x = 0: y = \cos(0) + \sqrt{3} \sin(0) = 1 \) - \( x = 2\pi: y = \cos(2\pi) + \sqrt{3} \sin(2\pi) = 1 \)- At critical points: - \( x = \frac{\pi}{3}: y = \cos(\frac{\pi}{3}) + \sqrt{3} \sin(\frac{\pi}{3}) = \sqrt{3} + \frac{1}{2} = \frac{2\sqrt{3} + 1}{2} \) - \( x = \frac{4\pi}{3}: y = \cos(\frac{4\pi}{3}) + \sqrt{3} \sin(\frac{4\pi}{3}) = -\sqrt{3} - \frac{1}{2} \)Absolute maximum is at \( x = \frac{\pi}{3} \) and absolute minimum is at \( x = \frac{4\pi}{3} \).

Step 4: Find the Second Derivative

To find inflection points, compute the second derivative: \[ y'' = -\cos{x} - \sqrt{3} \sin{x} \]

Step 5: Determine Inflection Points

Set the second derivative equal to zero to find potential inflection points: \[ -\cos{x} - \sqrt{3} \sin{x} = 0 \] Simplifying we find: \[ \tan{x} = -\frac{1}{\sqrt{3}} \] Solutions within the interval are \( x_1 = \frac{5\pi}{6}, \frac{11\pi}{6} \). These are the potential inflection points as sign changes occur on the second derivative around these points.

Key Concepts

Derivatives

In calculus, a derivative represents the rate of change of a function's value with respect to a change in its input. Essentially, it provides the slope of the function at any given point. For the function

• \( y = \cos{x} + \sqrt{3} \sin{x} \)

we calculate the first derivative:

• \( y' = -\sin{x} + \sqrt{3} \cos{x} \)

This derivative tells us how the function's slope behaves as \( x \) changes.
To determine extrema and critical points, setting this derivative equal to zero is crucial as it pinpoints where slopes flatten out, hinting at possible peak or trough points.
Moreover, finding the second derivative

• \( y'' = -\cos{x} - \sqrt{3} \sin{x} \)

helps analyze the shape of the graph, indicating concavity and potential inflection points. Derivatives, essentially, open the window to understanding the dynamic behavior of functions.

Critical Points

Critical points are essential in identifying the behavior of functions.
These points occur where the first derivative is either zero or undefined.
They show us where the function's slope varies—indicative of potential maximums, minimums, or even saddle points.
In the case of the function

• \( y = \cos{x} + \sqrt{3} \sin{x} \)

we simplify the equation

• \( -\sin{x} + \sqrt{3} \cos{x} = 0 \).

This provides

• \( \tan{x} = \sqrt{3} \)

leading to solutions within the given interval:

• \( x = \frac{\pi}{3} \) and \( x = \frac{4\pi}{3} \).

By evaluating our original function at these points and at endpoints, we confirm local and absolute extrema.
This understanding of critical points aids in graphing functions and analyzing their specific characteristics.

Inflection Points

Inflection points are fascinating as they indicate where a function changes its curvature—from concave upward to concave downward or vice versa.
Finding these points requires the second derivative of the function.For

• \( y = \cos{x} + \sqrt{3} \sin{x} \)

we have

• \( y'' = -\cos{x} - \sqrt{3} \sin{x} \)

Setting this second derivative to zero

• \( -\cos{x} - \sqrt{3} \sin{x} = 0 \)

allows us to find potential inflection points where the curve shifts its direction.
Simplifying

• \( \tan{x} = -\frac{1}{\sqrt{3}} \)

gives solutions

• \( x = \frac{5\pi}{6} \) and \( x = \frac{11\pi}{6} \).

These points reveal the dynamic nature of the function's graph, indicating where the rate of change in the slope itself changes.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially when dealing with periodically oscillating systems such as waves or rotations.
In

• \( y = \cos{x} + \sqrt{3} \sin{x} \)

these functions modulate the amplitude and phase of the function's output, dependent on the angle \( x \).
The derivative relationships to these functions are fundamental and provide critical insights:

• \( \frac{d}{dx} \cos{x} = -\sin{x} \)
• \( \frac{d}{dx} \sin{x} = \cos{x} \)

These derivatives assist in constructing important parts of laws and theories throughout mathematics and physics due to their repetition and predictable nature.
Understanding how these functions transform through calculus enables us to better conceptualize their roles in various applied fields, ensuring we can navigate through complex problems effectively.