Problem 70 - Algebra and Trigonometry

Question

\(67-70\). Find the maximum and minimum values of the function. $$ y=\frac{\cos x}{2+\sin x} $$

Step-by-Step Solution

Answer

The maximum value is \( \frac{\sqrt{3}}{3} \) and the minimum value is \( -\frac{\sqrt{3}}{3} \).

Step 1: Identify the Function and Domain

The function is given as \( y = \frac{\cos x}{2 + \sin x} \). Since the denominator cannot be zero, the domain is restricted to values where \( 2 + \sin x eq 0 \). The sine function has values between -1 and 1, making the denominator always positive, ranging from 1 to 3. Therefore, the function is defined for all values of \( x \).

Step 2: Find Critical Points

To find the critical points, differentiate the function with respect to \( x \). Using the quotient rule, the derivative is:\[ y' = \frac{(2 + \sin x)(-\sin x) - \cos x \cos x}{(2 + \sin x)^2} \]Simplifying yields:\[ y' = \frac{-2\sin x - \sin^2 x - \cos^2 x}{(2 + \sin x)^2} \]Since \( \sin^2 x + \cos^2 x = 1 \), this simplifies to:\[ y' = \frac{-2\sin x - 1}{(2 + \sin x)^2} \]Setting \( y' = 0 \) to find critical points gives:\[-2\sin x - 1 = 0 \Rightarrow \sin x = -\frac{1}{2} \]Solving this equation, \( x = \frac{7\pi}{6} + 2k\pi \) and \( x = \frac{11\pi}{6} + 2k\pi \), where \( k \) is an integer.

Step 3: Evaluate Function at Critical Points

Substitute the critical points into the original function to find their corresponding \( y \) values.- For \( x = \frac{7\pi}{6} \): \[ y = \frac{\cos \frac{7\pi}{6}}{2 + \sin \frac{7\pi}{6}} = \frac{-\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}} = \frac{-\sqrt{3}/2}{1.5} = -\frac{\sqrt{3}}{3} \]- For \( x = \frac{11\pi}{6} \): \[ y = \frac{\cos \frac{11\pi}{6}}{2 + \sin \frac{11\pi}{6}} = \frac{\frac{\sqrt{3}}{2}}{2 - \frac{1}{2}} = \frac{\sqrt{3}/2}{1.5} = \frac{\sqrt{3}}{3} \]

Step 4: Determine Boundary Behavior

Consider the behavior of \( y \) as \( \sin x \) approaches its extreme values. The maximum possible value in the denominator is 3 when \( \sin x = 1 \), yielding:\[ y = \frac{\cos x}{3} \quad \text{with range} \quad [-1/3, 1/3] \]Similarly, the minimum possible value in the denominator is 1 when \( \sin x = -1 \), giving:\[ y = \cos x \quad \text{with range} \quad [-1, 1] \]

Step 5: Conclude with Maximum and Minimum Values

The critical points yield \( y = -\frac{\sqrt{3}}{3} \) and \( \frac{\sqrt{3}}{3} \) as potential extremal values. The calculations with the extreme values of \( \sin x \) show that the maximum value for the function is \( 1 \) and the minimum value is \( -1 \). However, within the range of the critical points and boundary behavior, observe that the maximum value approached is \( \frac{\sqrt{3}}{3} \) and the minimum is \( -\frac{\sqrt{3}}{3} \). Hence, the function does not actually attain 1 or -1, but it is bounded accordingly.

Key Concepts

Critical Points

Critical points of a function are where its derivative equals zero or the derivative does not exist. These points are significant in identifying where a function may have maximum or minimum values, known as extrema. To find the critical points of the function \( y = \frac{\cos x}{2 + \sin x} \), we first differentiate it with respect to \( x \). The derivative, once simplified, equals \( y' = \frac{-2\sin x - 1}{(2 + \sin x)^2} \). By setting this derivative equal to zero \( (y' = 0) \), we solve for \( \sin x = -\frac{1}{2} \), giving us critical points at \( x = \frac{7\pi}{6} + 2k\pi \) and \( x = \frac{11\pi}{6} + 2k\pi \) where \( k \) is an integer.
These points are potential locations for extrema since changes in direction of the function often occur here.

Differentiation using Quotient Rule

The quotient rule in calculus is a formula used to differentiate a function that is the quotient of two other functions. If we have a function \( y = \frac{u(x)}{v(x)} \), then its derivative is given by the rule:

• \( y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \)

For our function \( y = \frac{\cos x}{2 + \sin x} \), we set \( u(x) = \cos x \) and \( v(x) = 2 + \sin x \).
Applying the quotient rule:

• \( u'(x) = -\sin x \) (derivative of cosine)
• \( v'(x) = \cos x \) (derivative of the sine contributing to the denominator)

Substitute these into the quotient rule to find that \( y' = \frac{(2 + \sin x)(-\sin x) - \cos x \cdot \cos x}{(2 + \sin x)^2} \).
This operation is important as it reveals the behavior of the original function through its slope or rate of change.

Behavior of Sine Function

Understanding the behavior of the sine function is crucial since it appears in both the numerator and denominator in our function \( y = \frac{\cos x}{2 + \sin x} \). The sine function, \( \sin x \), oscillates between -1 and 1.
This oscillating nature means that the denominator of our function, \( 2 + \sin x \), will vary between 1 and 3 since:

• Minimum: \( 2 + (-1) = 1 \)
• Maximum: \( 2 + 1 = 3 \)

When solving for critical points, \( \sin x = -\frac{1}{2} \) corresponds to individual angles on the unit circle where the sine function reaches this specific value. This periodic behavior is why critical points appear at intervals of \( \pi \) radians apart for sine.

Bounded Function Behavior

Functions can be bounded, meaning they have maximum and minimum possible values within a specific interval. For the function \( y = \frac{\cos x}{2 + \sin x} \), since \( 2 + \sin x \) is always positive, \( y \) is well-defined for all \( x \).
Considering the boundary behaviors:

• Maximum value of denominator, \( 3 \), makes the function's range narrower: \( -\frac{1}{3} \) to \( \frac{1}{3} \).
• Minimum value of denominator, \( 1 \), allows the widest range, i.e., from \( -1 \) to \( 1 \).

This bounded nature indicates that the function cannot reach values beyond these limitations. Critical points provide local extrema that lie between these bounds. Therefore, while a critical point provides \( y = \frac{\sqrt{3}}{3} \) or \( -\frac{\sqrt{3}}{3} \), it upholds the bounded characteristic of the function which goes up to \( \pm 1 \).
Understanding this helps to clarify what possible values the function can genuinely attain.