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Chapter 5: Problem 33

Find all local maximum and minimum points by the second derivative test. $$ y=\cos ^{2} x-\sin ^{2} x $$

Short Answer

Expert verified
Local minima at \( x = 2k\pi \); local maxima at \( x = (2k+1)\pi/2 \).

Step by step solution

01

Find the First Derivative

To find local maximum and minimum points, we start by finding the first derivative of the function. Given the function \( y = \cos^2 x - \sin^2 x \), apply the chain rule to differentiate: \( y' = -2\cos(x)\sin(x) + 2\sin(x)\cos(x) = 0 \). Simplify this: \( y' = 0 \).
02

Simplify the First Derivative Equation

The first derivative \( y' = -2\cos(x)\sin(x) + 2\sin(x)\cos(x) = 0 \) simplifies to \( y' = 0 \). Using the identity \( a - a = 0 \), leads to \( 0 = 0 \), which does not help. Thus, let鈥檚 factor it differently: \( y' = 2\sin(x)\cos(x) \). Simplifying further, \( y' = \sin(2x) = 0 \).
03

Find the Critical Points

Solve \( \sin(2x) = 0 \). This equation holds true when \( 2x = n\pi \) where \( n \) is an integer. Solving for \( x \), we get \( x = \frac{n\pi}{2} \). These are our critical points.
04

Find the Second Derivative

Take the second derivative of \( y \). Starting from \( y' = \sin(2x) \): use the derivative \( y'' = \frac{d}{dx} \sin(2x) = 2\cos(2x) \).
05

Apply the Second Derivative Test

To determine if each critical point is a local maximum or minimum, substitute \( x = \frac{n\pi}{2} \) into the second derivative \( y'' = 2\cos(2x) \). For \( x = \frac{n\pi}{2} \), this becomes \( y'' = 2\cos\left(n\pi\right) \). If \( n \) is even, \( y'' = 2 \times 1 = 2 \) (local minimum). If \( n \) is odd, \( y'' = 2 \times (-1) = -2 \) (local maximum).
06

Conclude the Local Extrema

Thus, the local minima occur at \( x = 2k\pi \) and the local maxima occur at \( x = (2k+1)\frac{\pi}{2} \), where \( k \) is any integer.

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Key Concepts

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

Local Maximum
In calculus, a local maximum of a function is a point where the function value is greater than its neighboring values. Think of a local maximum as the top of a hill within a certain area. To find a local maximum using the second derivative test, you follow these steps:
  • First, locate the critical points by setting the first derivative equal to zero and solving for the variable.
  • Next, calculate the second derivative at each critical point.
  • If the second derivative is negative (\( y'' < 0 \)), the function curves downwards, indicating a local maximum.
For the given function \( y = \cos^2 x - \sin^2 x \), critical points occur at \( x = \frac{n\pi}{2} \) where \( n \) is an integer. From the solution, we see that local maxima appear when \( n \) is odd, because \( y'' = -2 \) is negative, confirming the downward concavity.
Local Minimum
Local minima are points where a function is lower than the neighboring values. Picture it as the bottom of a valley on a graph. Here's how you use the second derivative test to identify local minima:
  • Identify critical points by finding where the first derivative equals zero.
  • Evaluate the second derivative at these critical points.
  • If the second derivative is positive (\( y'' > 0 \)), the graph curves upwards, indicating a local minimum.
Using the second derivative method for our function, we concluded that when \( n \) is even, \( y'' = 2 \), which is positive. Therefore, the function reaches a local minimum at these critical points. So, local minima occur at \( x = 2k\pi \).
Critical Points
Critical points are values in the domain of a function where the derivative is zero or undefined. They are essential for identifying potential local maxima or minima:
  • First, compute the first derivative of the function.
  • Find where this derivative is zero or fails to exist.
  • In the exercise, the function \( y' = \sin(2x) \) became zero at certain points. This occurs when \( 2x = n\pi \), giving us \( x = \frac{n\pi}{2} \)
These \( x \) values are the critical points where each potential extremum can be examined with the second derivative.
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. Composite functions are functions nested inside one another. The chain rule formula is:\[ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \]This means you differentiate the outer function and then multiply it by the derivative of the inner function.In our exercise, we applied the chain rule to differentiate \( y = \cos^2 x - \sin^2 x \). Since each term involves applying the trigonometric chain rule:
  • For \( \cos^2 x \), let \( u = \cos x \), which gives \( u^2 \). The derivative of this is \( 2u \cdot (-\sin x) \ = -2\cos x \sin x \).
  • For \( \sin^2 x \), let \( v = \sin x \), which gives \( v^2 \). The derivative becomes \( 2v \cdot (\cos x) = 2\sin x \cos x \).
The chain rule helps break down these components into simpler parts for differentiation.

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Most popular questions from this chapter

Let \(n\) be an integer greater than or equal to two, and suppose \(f\) is a polynomial of degree n. How many inflection points can \(f\) have? Hint: Use the second derivative test and the fundamental theorem of algebra.

Let \(f(\theta)=\cos ^{2}(\theta)-2 \sin (\theta) .\) Find the intervals where \(f\) is increasing and the intervals where \(f\) is decreasing in \([0,2 \pi] .\) Use this information to classify the critical points of \(f\) as either local maximums, local minimums, or neither.

Sketch the curves. Identify clearly any interesting features, including local maximum and minimum points, inflection points, asymptotes, and intercepts. $$ y=\left(e^{x}+e^{-x}\right) / 2 $$

Find all critical points and identify them as local maximum points, local minimum points, or neither. $$ y=x^{3}-9 x^{2}+24 x $$

Describe the concavity of the functions below. $$ y=\sin ^{3} x $$
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