Problem 94 - Calculus Single Variable

Question

In each of Exercises \(91-94\), calculate and plot the derivative \(f^{\prime}\) of the given function \(f\). Use this plot to identify candidates for the local extrema of \(f\). Add the plot of \(f\) to the window containing the graph of \(f^{\prime} .\) From this second plot, determine the behavior of \(f\) at each candidate for a local extremum. $$ f(x)=\sin ^{2}(x)-x^{3}+5 x+20 $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \sin(2x) - 3x^2 + 5 \). Plot both the function and its derivative to identify and classify local extrema.

Step 1: Find the Derivative

To find the derivative of the function \( f(x) = \sin^2(x) - x^3 + 5x + 20 \), apply the chain rule and power rule. The derivative is calculated as follows:1. Differentiate \( \sin^2(x) \): Apply the chain rule to get \( 2\sin(x) \cos(x) = \sin(2x) \).2. Differentiate \( -x^3 \): The power rule gives us \( -3x^2 \).3. Differentiate \( 5x \): This results in \( 5 \).4. The derivative of the constant \( 20 \) is \( 0 \).Therefore, the derivative \( f'(x) = \sin(2x) - 3x^2 + 5 \).

Step 2: Plot the Derivative

Plot the function \( f'(x) = \sin(2x) - 3x^2 + 5 \) over an appropriate interval. This will provide insights into where the function's slope is zero, indicating potential local maxima or minima.

Step 3: Identify Critical Points

Critical points are where the derivative \( f'(x) = 0 \). Solve the equation:\[ \sin(2x) - 3x^2 + 5 = 0 \]This may require numerical methods or graphing technology to approximate the solutions. These points are where \( f(x) \) may have local extrema.

Step 4: Plot the Original Function

Now, plot the original function \( f(x) = \sin^2(x) - x^3 + 5x + 20 \) along with the derivative plot. Overlay the graphs to visualize where \( f(x) \) might have local minima or maxima by checking the behavior change in \( f'(x) \).

Step 5: Analyze Behavior at Extrema

At each critical point identified, examine the plot to determine if these points are local maxima, minima, or points of inflection. A change from positive to negative in \( f'(x) \) suggests a local maximum, while a change from negative to positive suggests a local minimum.

Key Concepts

Critical Points

Critical points are very important in understanding the behavior of a function. These are the points where the derivative of the function equals zero or is undefined. For a continuous function, critical points can provide potential locations where the function has local maxima, minima, or saddle points. In other words, they help us identify where the slope of the tangent to the function changes direction.
To find these critical points, we set the derivative equal to zero and solve for the variable. In the exercise, the derivative is given by:

• \[ f'(x) = \sin(2x) - 3x^2 + 5 \]

Solving this equation may require graphical or numerical methods, as it might be intricate to solve algebraically. Once found, these critical points can then be further analyzed to determine the nature of the point in terms of local maximums, minimums, or points of inflection.

Local Extrema

Local extrema refer to the local maximums or minimums of a function. A function can have a local maximum where it reaches the highest point in a small interval, and a local minimum where it reaches the lowest. By examining the critical points, we can identify these local extrema.
Once the critical points are identified from the derivative, the behavior of the function at these points should be tested. Typically, we use the first or second derivative test to determine the nature of these extrema. Consider this:

• If the derivative changes from positive to negative at a point, it indicates a local maximum.
• If the derivative changes from negative to positive, it points to a local minimum.

Visualizing the function and its derivative together can significantly help understand the behavior at these critical points, allowing you to confirm if they are local maxima or minima.

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. It’s essential when dealing with functions where one function is inside another. If you have a function such as \( f(g(x)) \), the chain rule helps us find its derivative. It states that the derivative of \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
In the exercise, the chain rule is used to differentiate the term \( \sin^2(x) \). Treat \( \sin(x) \) as an inner function and \( x^2 \) as an outer function. The chain rule gives us:

• \[ \frac{d}{dx} (\sin^2(x)) = 2 \sin(x) \cos(x) = \sin(2x) \]

This technique is crucial, especially when editing complex functions correctly without getting bogged down by errors.

Power Rule

The Power Rule is one of the simplest and most commonly used techniques for differentiation. It is used when you need to find the derivative of a function with a term involving a power of \( x \).
The general formula states that for a term \( x^n \), the derivative is \( nx^{n-1} \). This straightforwardly lets us differentiate polynomials and other expressions containing power terms.
For the given exercise, the power rule is applied to the term \(-x^3\), resulting in the derivative \(-3x^2\). The application is simple:

• Drop the exponent down as a constant pre-multiplier (3 in this case).
• Reduce the exponent by one to find the resulting derivative term (-3x²).

Understanding this rule is essential as it paves the way for more advanced derivative calculations and is one of the first steps in grasping calculus basics.