Problem 49 - University Calculus: Early Transcendentals

Question

a. Find the derivative \(f^{\prime}(x)\) of the given function \(y=f(x)\) b. Graph \(y=f(x)\) and \(y=f^{\prime}(x)\) side by side using separate sets of coordinate axes, and answer the following questions. c. For what values of \(x\), if any, is \(f^{\prime}\) positive? Zero? Negative? d. Over what intervals of \(x\) -values, if any, does the function \(y=f(x)\) increase as \(x\) increases? Decrease as \(x\) increases? How is this related to what you found in part (c)? (We will say more about this relationship in Section \(4.3 .\) ) $$y=-x^{2}$$

Step-by-Step Solution

Answer

a. The derivative is \( f'(x) = -2x \). c. Positive for \( x < 0 \), zero at \( x = 0 \), and negative for \( x > 0 \). d. Increase for \( x < 0 \), decrease for \( x > 0 \).

Step 1: Differentiate the Function

To find the derivative of the function, we start with the given function: \[ y = f(x) = -x^2 \]The derivative, using the power rule \( \frac{d}{dx}x^n = nx^{n-1} \), becomes:\[ f'(x) = \frac{d}{dx}(-x^2) = -2x \]

Step 2: Graphing the Functions

Now, let's consider the graphs:1. **Graph of \( y = -x^2 \):** This is a downward-opening parabola with vertex at the origin (0,0).2. **Graph of \( y = -2x \):** This is a straight line with a negative slope of -2, meaning it goes downward from left to right.Draw these graphs side by side on separate coordinate axes to visualize the relationship.

Step 3: Determine Sign of Derivative

We need to explore where the derivative \( f'(x) = -2x \) is positive, zero, or negative:- \( f'(x) > 0 \) when \( -2x > 0 \), which implies \( x < 0 \).- \( f'(x) = 0 \) when \( -2x = 0 \), which implies \( x = 0 \).- \( f'(x) < 0 \) when \( -2x < 0 \), which implies \( x > 0 \).

Step 4: Determine Intervals of Increase and Decrease

To determine where the function increases or decreases, we look at the sign of \( f'(x) \):- The function \( f(x) = -x^2 \) increases for \( x < 0 \) because \( f'(x) > 0 \).- The function decreases for \( x > 0 \) because \( f'(x) < 0 \).- At \( x = 0 \), the function neither increases nor decreases as this is the vertex.This analysis shows that the intervals where \( f(x) \) increases correspond to where \( f'(x) \) is positive and decreases where \( f'(x) \) is negative.

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus used for differentiating functions of the form \(x^n\), where \(n\) is a real number. To apply this rule, you simply multiply the exponent \(n\) by the coefficient in front of \(x\) and then reduce the exponent by one. This is expressed as:

• \( \frac{d}{dx}x^n = nx^{n-1} \)

In our exercise, the function given is \(y = -x^2\). Using the power rule, we differentiate and get the derivative \(f'(x) = -2x\). This is because we multiply the exponent 2 by the coefficient -1, resulting in -2, and reduce the exponent by one, changing \(x^2\) to \(x^1\), which is simply \(x\). This simple calculation is powerful enough to uncover the slope of the function at any given point on its graph.

Graphing Functions

Graphing functions helps visualize the behavior of mathematical equations in a way that numeric or algebraic representations alone cannot. In our case, we graph two functions:

• \(y = -x^2\)
• \(y = -2x\)

The graph of \(y = -x^2\) is a parabola opening downwards with its vertex at the origin (0,0). This shape is typical for quadratic equations with a negative leading coefficient.
Meanwhile, \(y = -2x\) is a straight line with a negative slope of -2, indicating that as \(x\) increases, \(y\) decreases. Visualizing these graphs together highlights the relationship between a function and its derivative—where the line \(y = -2x\) represents the rate of change (slope) of the parabola. Understanding how these graphs relate aids in predicting the function's growth or decline over specific intervals.

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing is key in analyzing its behavior. A function \(f(x)\) is increasing on an interval if, as \(x\) increases within this interval, \(f(x)\) also increases. Conversely, it is decreasing on an interval if \(f(x)\) decreases as \(x\) increases.
For our function \(f(x) = -x^2\), we use its derivative \(f'(x) = -2x\) to determine these intervals.

• Where \(f'(x) > 0\), the function is increasing. This occurs for \(x < 0\).
• Where \(f'(x) < 0\), the function is decreasing. This happens for \(x > 0\).

At \(x = 0\), the function is neither increasing nor decreasing because this is where it reaches its maximum height, known as the vertex. Here, the graph flattens out before changing direction.

Sign of the Derivative

The sign of a derivative is a significant indicator of a function's behavior. It tells us whether the function is increasing, decreasing, or stable at a particular point.

• If \(f'(x) > 0\), the function is increasing at \(x\).
• If \(f'(x) < 0\), the function is decreasing at \(x\).
• If \(f'(x) = 0\), the function is stable or constant at that point; this often signifies a local maximum or minimum.

For \(f'(x) = -2x\), the derivative is:

• Positive for \(x < 0\)
• Zero for \(x = 0\)
• Negative for \(x > 0\)

These signs guide us in determining the nature of the function \(f(x) = -x^2\) over its domain. Such analysis helps in understanding the dynamic nature of calculus functions and how their graphs behave.