Problem 33 - Thomas Calculus

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. \begin{equation} y=2 x-3 x^{2 / 3} \end{equation}

Step-by-Step Solution

Answer

Local minimum at x=1 with no inflection points.

Step 1: Find the first derivative

To identify critical points where local extrema might occur, we first find the derivative of the function. Given the function \( y = 2x - 3x^{2/3} \), apply the power rule. The derivative is:\[ y' = \frac{d}{dx}(2x) - \frac{d}{dx}(3x^{2/3}) = 2 - 2x^{-1/3}. \]

Step 2: Solve for critical points

Set the derivative equal to zero to find critical points: \[ 2 - 2x^{-1/3} = 0. \] Solve for \( x \): \[ 2 = 2x^{-1/3} \] \[ x^{-1/3} = 1 \] \[ x = 1. \] The critical point is \( x = 1 \).

Step 3: Determine the second derivative

To find inflection points and analyze the concavity, calculate the second derivative. Differentiate \( y' = 2 - 2x^{-1/3} \) again to obtain:\[ y'' = \frac{d}{dx}(2 - 2x^{-1/3}) = \frac{2}{3}x^{-4/3}. \]

Step 4: Solve for inflection points

Set the second derivative to zero to find any inflection points:\[ \frac{2}{3}x^{-4/3} = 0. \] Since the expression \( x^{-4/3} \) does not equal zero for any real \( x \), there are no inflection points.

Step 5: Analyze for concavity and extrema

For \( x > 1 \), \( y'' > 0 \), indicating the function is concave up. For \( x < 1 \), \( y'' < 0 \), indicating concave down. Thus, the critical point at \( x = 1 \) is a local minimum.

Step 6: Evaluate at endpoints for absolute extrema

Since there are no endpoints specified by an interval, evaluate behavior as \( x \to \infty \) and \( x \to -\infty \). As \( x \to \infty \), the function tends towards infinity. As \( x \to -\infty \), the function will go towards negative infinity. Thus, no absolute extrema.

Step 7: Graph the function

Plot the function \( y = 2x - 3x^{2/3} \). Mark the local minimum at \( (1, y(1)) \). Note the concavity changes at the critical point, showing a 'valley' for the local minimum.

Key Concepts

First Derivative

The first derivative of a function helps us understand the rate at which the function changes. When we calculate the first derivative, we apply differentiation rules to find how the output value of the function changes as the input value varies. In this exercise, we start with the function \( y = 2x - 3x^{2/3} \). By applying the basic power rule of differentiation, the first derivative \( y' \) becomes \( 2 - 2x^{-1/3} \).
This derivative expression indicates the slope of the tangent line to the curve at any given point. A positive derivative suggests the function is increasing, while a negative derivative indicates it's decreasing. Finding where the first derivative equals zero or doesn't exist helps us locate the critical points, which are potential locations for local extrema.

Critical Points

Critical points are specific points on the graph of a function where the first derivative is zero or undefined. These points are important because they are candidates for where the function might have a local maximum or minimum. To find the critical points in our function, we first solve the equation \( 2 - 2x^{-1/3} = 0 \).
After solving this equation, we identify \( x = 1 \) as the only critical point. Critical points must be evaluated further through the second derivative test or other means to confirm whether they are indeed local extrema. It's not enough just to find where the slope is zero; we need more information about the behavior of the function around these points.

Second Derivative

The second derivative of a function provides information about the curvature or concavity of the graph. By differentiating the first derivative of \( y' = 2 - 2x^{-1/3} \) again, we find the second derivative \( y'' = \frac{2}{3}x^{-4/3} \). The second derivative helps us determine how the slope is changing.

• If \( y'' > 0 \) at a point, the function is concave up there, resembling a 'U' shape.
• If \( y'' < 0 \), the function is concave down, forming an upside-down 'U'.

This information is essential to evaluate how the function behaves around its critical points. The second derivative is also used to identify inflection points where the concavity changes, although in our case there are no real \( x \) values making \( y'' = 0 \).

Concavity

Concavity describes how a function curves and whether its graph opens upwards or downwards.
With the second derivative \( y'' = \frac{2}{3}x^{-4/3} \), we examine the concavity for different portions of the domain:

• For \( x > 1 \): \( y'' > 0 \), indicating that the function is concave up.
• For \( x < 1 \): \( y'' < 0 \), indicating that the function is concave down.

This analysis explains that the curve changes its curvature at the critical point \( x = 1 \). In this case, it moves from being concave down to concave up, suggesting a valley at that point. Understanding concavity is vital for sketching accurate graphs and predicting the nature of extremum points.

Local Extrema

Local extrema refer to the highest or lowest points within a particular region of a function's graph. These include local maxima and minima. From the earlier analysis,

• At \( x = 1 \), there is a local minimum because the concavity changes from downwards to upwards as \( y'' \) moves from negative to positive.

However, our examination of the function's behavior at the endpoints as \( x \to \infty \) and \( x \to -\infty \) showed that there are no absolute extrema since the function tends towards infinity in both directions.
In a graphical representation, local extrema manifest as "peaks" and "valleys" within a limited view, providing critical information about the function's landscape. Identifying these points is a core part of understanding and analyzing functions in calculus.