Problem 22 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ k(x)=\frac{\cos x}{1+\sin x} $$

Step-by-Step Solution

Answer

The function has a relative minimum at \( x = \frac{3\pi}{2} \).

Step 1: Find the Derivative

To apply the First Derivative Test, we first need to find the derivative of the function \( k(x) = \frac{\cos x}{1+\sin x} \). Use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = \cos x \) and \( v = 1 + \sin x \). Thus, \( u' = -\sin x \) and \( v' = \cos x \). Applying the quotient rule gives: \[ k'(x) = \frac{-\sin x (1+\sin x) - \cos x \cos x}{(1+\sin x)^2} = \frac{-\sin x - \sin^2 x - \cos^2 x}{(1+\sin x)^2} \] Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to: \[ k'(x) = \frac{-\sin x - 1}{(1+\sin x)^2} \] This is the first derivative of the function.

Step 2: Determine Critical Points

Critical points occur where the derivative is either zero or undefined. Since the denominator \((1+\sin x)^2\) must be nonzero, critical points occur where the numerator \(-\sin x - 1 = 0\). Solving for \(x\) gives: \[ -\sin x - 1 = 0 \] \[ \sin x = -1 \] The solution to \( \sin x = -1 \) is \( x = \frac{3\pi}{2} + 2k\pi \), where \( k \) is an integer.

Step 3: Test Intervals Around Critical Points

To use the First Derivative Test, examine values around the critical points. Consider a number slightly less than and greater than \( \frac{3\pi}{2} \) (like \( x = \frac{3\pi}{2} - \epsilon \) and \( x = \frac{3\pi}{2} + \epsilon \)). The signs of \( k'(x) \) at these points determine increasing or decreasing behavior:- For \( x \in (\pi, \frac{3\pi}{2}) \), pick \( x = \frac{5\pi}{4} \). \( \sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} \) and thus \( k'(x) = \frac{-(-\frac{\sqrt{2}}{2}) - 1}{(1-\frac{\sqrt{2}}{2})^2} < 0 \). Function is decreasing.- For \( x \in (\frac{3\pi}{2}, 2\pi) \), pick \( x = \frac{7\pi}{4} \). \( \sin(\frac{7\pi}{4}) = -\frac{\sqrt{2}}{2} \) and thus \( k'(x) = \frac{-(-\frac{\sqrt{2}}{2}) - 1}{(1-\frac{\sqrt{2}}{2})^2} > 0 \). Function is increasing.

Step 4: Conclusion About Relative Extrema

Since the function changes from decreasing to increasing around \( x = \frac{3\pi}{2} \), there is a relative minimum at this point. Use the First Derivative Test which states that if \( k'(x) \) changes from negative to positive at a point, then \( x \) is a relative minimum.

Key Concepts

Quotient Rule

The quotient rule is a technique used in calculus to find the derivative of the ratio of two functions. When you have a function that is written as a quotient, such as \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule applies. The formula for the quotient rule is:

• \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)

You begin by identifying \( u(x) \) and \( v(x) \) from the function. Then, calculate their derivatives, \( u'(x) \) and \( v'(x) \). The result of the quotient rule provides the derivative of the original function. This rule is particularly useful when dealing with expressions where two functions are divided. By following this method, you ensure an accurate calculation of the derivative, necessary for applying further calculus concepts like finding extrema.

Derivative Calculus

Derivative calculus is all about understanding how functions change. The derivative of a function represents the rate of change of the function’s value with respect to a change in the input value. For a function \( f(x) \), its derivative, denoted as \( f'(x) \) or \( \frac{df}{dx} \), helps us understand how the function behaves under small changes of \( x \).

• It tells us the slope of the tangent line to the curve at any point.
• It signifies how fast a function is increasing or decreasing at a particular point.

Taking derivatives is fundamental in calculus for analyzing functions. With derivatives, we can identify important properties of graphs such as critical points, which are locations where the function’s rate of change is zero or undefined. This is a crucial step in determining extrema and analyzing the behavior of functions.

Relative Extrema

Relative extrema are points on a graph where a function reaches local minimum or maximum values. The First Derivative Test is a handy tool used in calculus to determine these points. To apply this test, you first find the derivative of the function. Then, determine the critical points by setting the derivative equal to zero and solving for \( x \). These are the points where potential relative minima or maxima can occur.Next, examine intervals around the critical points to see how the function’s derivative behaves. If the derivative changes from negative to positive, it indicates a relative minimum. Conversely, if it changes from positive to negative, it suggests a relative maximum.Understanding relative extrema is valuable because it gives insights into the overall shape of the function, allowing you to predict and describe the function's behavior more comprehensively. These points often correspond to peak or trough points on a graph, which are significant for understanding the nature of the function.