Problem 42 - Calculus of a Single Variable

Question

Using the Second Derivative Test In Exercises \(31-42\) , find all relative extrema. Use the Second Derivative Test where applicable. $$ f(x)=2 \sin x+\cos 2 x, \quad[0,2 \pi] $$

Step-by-Step Solution

Answer

By applying the second derivative test, it is found that the function \(f(x)=2\sin{x}+\cos{2x}\) has a local maximum at \(x=\pi/2\) and a local minimum at \(x=3\pi/2\).

Step 1: Function Derivative

Firstly, differentiate the given function \(f(x)=2\sin{x}+\cos{2x}\) to get \(f'(x)=2\cos{x}-2\sin{2x}\). This derivative will be used to find the critical points.

Step 2: Critical Points

To find the critical points, solve the equation \(f'(x) = 0\). That is, solve \(2\cos{x}-2\sin{2x}=0\). Upon solving, you'll get two critical points as \(x=\pi/2\) and \(x=3\pi/2\).

Step 3: Second Derivative

Now, to perform the Second Derivative Test, the second derivative of the function must be calculated. Differentiating \(f'(x)\) will yield \(f''(x)=-2\sin{x}-4\cos{2x}\).

Step 4: Second Derivative Test

Substitute the critical points into the second derivative function: \(f''(\pi/2)=-2\sin{\pi/2}-4\cos{\pi}\) and \(f''(3\pi/2)=-2\sin{3\pi/2}-4\cos{3\pi}\). The conclusion can be achieved by noting that if the second derivative at a point is positive, it is a local minimum, and if it is negative, it is a local maximum. At \(x=\pi/2\), the second derivative is negative, indicating that it is a local maximum point, while at \(x=\pi/2\) it is positive, indicating that it is a local minimum point.

Key Concepts

Critical Points

Critical points play a key role when examining the behavior of a function. They are the values of the variable where the function's first derivative is zero or undefined. To determine critical points:

• Begin by finding the derivative of the function.
• Solve the equation where this derivative equals zero ( \( f'(x) = 0 \) ) to find critical points.

Identifying these points will help in understanding where a function might have local maxima, minima, or points of inflection. In our exercise, solving \( 2\cos{x} - 2\sin{2x} = 0 \) gives us the critical points \( x = \pi/2 \) and \( x = 3\pi/2 \). These points are where the slope of the tangent to the function is zero.

Relative Extrema

Locating the relative extrema of a function involves understanding where the function reaches local maxima or minima. These points occur at the critical points you have calculated. To confirm whether a critical point is a maximum or a minimum, the Second Derivative Test is employed. This test requires:

• Calculating the second derivative of the function.
• Substituting the critical points into this second derivative.

If the second derivative at a critical point is positive, this point is a local minimum. If it's negative, it's a local maximum. In our case, at \( x = \pi/2 \), the second derivative is negative, pointing to a local maximum, while at \( x = 3\pi/2 \), the positive value indicates a local minimum.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in various fields, including calculus. They are periodic functions, meaning they repeat at regular intervals, which significantly affects their derivatives and critical points.

• The function \( \sin(x) \) has derivatives that cycle through \( \cos(x), -\sin(x), -\cos(x), \sin(x) \) again.
• Similarly, \( \cos(2x) \) involves adjusting the period and amplitude due to the multiplier 2.

In this exercise, we worked with \( \sin(x) \) and \( \cos(2x) \), examining their combined effects. Knowing their traits helps solve complex calculus problems that involve periodic behavior, like the identification of critical points and extrema.

Differentiation

Differentiation is a fundamental process in calculus used to find the rate at which a function is changing. It helps in determining the slope at any given point and identifying critical points, which is why it is crucial in our exercise.

• The first derivative \( f'(x) \) gives the slope of the function \( f(x) \).
• The second derivative \( f''(x) \) gives us the concavity, crucial for the Second Derivative Test.

For the function provided, differentiating \( f(x) = 2\sin{x} + \cos{2x} \) gave us \( f'(x) = 2\cos{x} - 2\sin{2x} \). Differentiating again resulted in \( f''(x) = -2\sin{x} - 4\cos{2x} \), essential for determining the nature of the critical points. Differentiation thus provides the stepping stones to resolving the entire problem.