Problem 17 - Calculus: Early Transcendentals

Question

Find all local maximum and minimum points by the second derivative test. $$ y=x^{2}-x $$

Step-by-Step Solution

Answer

The local minimum is at \( (\frac{1}{2}, -\frac{1}{4}) \); there's no local maximum.

Step 1: Find the first derivative

To begin identifying critical points, calculate the first derivative of the function \( y = x^2 - x \). The first derivative, \( y' \), is obtained by differentiating with respect to \( x \): \( y' = 2x - 1 \).

Step 2: Find critical points

Set the first derivative equal to zero to find the critical points: \( 2x - 1 = 0 \). Solving for \( x \) gives \( x = \frac{1}{2} \). This is a potential location for a local maximum or minimum.

Step 3: Find the second derivative

Calculate the second derivative to apply the second derivative test. The second derivative of \( y = x^2 - x \) is \( y'' = 2 \).

Step 4: Apply the second derivative test

Evaluate the second derivative at the critical point \( x = \frac{1}{2} \). Since \( y''(x) = 2 \) is positive, the function has a local minimum at \( x = \frac{1}{2} \).

Step 5: Determine the local minimum value

Substitute \( x = \frac{1}{2} \) into the original function to find the local minimum value: \( y = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \). So, the local minimum point is \( \left( \frac{1}{2}, -\frac{1}{4} \right) \).

Key Concepts

First Derivative and Its Role

The first derivative of a function is a fundamental concept in calculus. It represents the rate at which the function's value is changing at any given point. When you differentiate a function, you are essentially finding how the function behaves or changes as the input changes. In the context of our exercise with the function \( y = x^2 - x \), we compute the first derivative to be \( y' = 2x - 1 \).

• When \( y' > 0 \), the function is increasing.
• When \( y' < 0 \), the function is decreasing.
• When \( y' = 0 \), we have a potential critical point.

Finding where the first derivative equals zero is crucial because it helps locate critical points, which are the potential site of local maxima or minima. In our example, setting \( y' = 0 \) provides us with \( x = \frac{1}{2} \), indicating that this is a critical point where further analysis is needed.

The Second Derivative and the Test

The second derivative provides insight into the concavity of the function: whether the graph is curving upwards or downwards. This is essential to determine the nature of the critical points found using the first derivative. For the function \( y = x^2 - x \), the second derivative is \( y'' = 2 \).

• If \( y'' > 0 \), the function is concave up, suggesting a local minimum.
• If \( y'' < 0 \), the function is concave down, suggesting a local maximum.

Applying the second derivative test at the critical point \( x = \frac{1}{2} \) tells us \( y'' = 2 \), which is positive. This means the curve is concave upward at \( x = \frac{1}{2} \), indicating a local minimum. Therefore, the second derivative test confirms that we have a local minimum at this critical point.

Understanding Critical Points

Critical points are the places on a curve where the function's rate of change is zero, identified by setting the first derivative to zero. These points are pivotal in finding local extremes such as maxima or minima. In this exercise, \( x = \frac{1}{2} \) is established as a critical point.
However, not all critical points result in local maxima or minima. After pinpointing these points, we need further evaluation to ascertain their nature using the second derivative.
Ultimately, knowing the location and type of critical points helps in understanding the behavior of the function's graph—where it reaches peaks, troughs, or points of inflection. Addressing these points liberates us to find values like \( (\frac{1}{2}, -\frac{1}{4}) \), representing the local minimum of the function.