Problem 30 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{2 / 5}$$

Step-by-Step Solution

Answer

The local and absolute minimum is at \( (0,0) \). There are no inflection points.

Step 1: Find the Derivative

First, we need to find the first derivative of the function to identify any critical points where local extrema could occur. The given function is \( y = x^{2/5} \). We differentiate it with respect to \( x \):\[\frac{dy}{dx} = \frac{d}{dx}(x^{2/5}) = \frac{2}{5}x^{-3/5} = \frac{2}{5} \cdot \frac{1}{x^{3/5}}.\]

Step 2: Find Critical Points

To find critical points, we set the first derivative equal to zero or identify points where it is undefined. The derivative \( \frac{2}{5}x^{-3/5} \) is undefined at \( x = 0 \) because division by zero occurs. We also note that the derivative never reaches zero for other values of \( x \). Thus, the critical point is at \( x = 0 \).

Step 3: Evaluate the Function at Critical Points

We evaluate the given function at the critical point \( x = 0 \) to find the corresponding \( y \)-coordinate:\[y = (0)^{2/5} = 0.\]The function value at the critical point is \( y = 0 \). So the point \( (0,0) \) is a candidate for a local extremum.

Step 4: Classify the Critical Points

Since the derivative goes from being undefined to positive as \( x \) moves away from zero (for example, it is positive for positive \( x \) and approaches from negative increasing for negative \( x \)), \( x=0 \) is a local minimum. The absolute minimum is also at \( (0,0) \) since the function is non-negative everywhere.

Step 5: Find the Second Derivative

To determine inflection points, we find the second derivative:\[\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{2}{5}x^{-3/5}\right) = \frac{2}{5} \times \left(-\frac{3}{5}\right)x^{-8/5} = -\frac{6}{25}x^{-8/5}.\]

Step 6: Identify Inflection Points

Inflection points occur where the second derivative changes sign. However, \( \frac{d^2y}{dx^2} = -\frac{6}{25}x^{-8/5} \) is undefined at \( x=0 \) and does not change sign around \( x=0 \) since it remains negative for all \( x eq 0 \). Therefore, there are no inflection points.

Step 7: Graph the Function

The graph of \( y = x^{2/5} \) shows a vertical tangent at \( x=0 \) and is continuous everywhere else, with a gentle upward slope increasing away from 0 in both directions. It is concave down everywhere since the second derivative is negative for all \( x eq 0 \).

Key Concepts

Derivative

In calculus, a derivative represents the rate of change of a function. It tells us how a function value changes as its input changes slightly. For the function \( y = x^{2/5} \), finding the derivative involves applying the power rule for differentiation.

• The power rule states that if you have \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
• For our function \( y = x^{2/5} \), the derivative is \( \frac{dy}{dx} = \frac{2}{5}x^{-3/5} \).

This derivative helps us identify how the function's graph is behaving at any point, such as determining where it might have peaks, valleys, or flat points. It's like having a map showing steep paths and flat roads on a hiking trail. The derivative also lets us find critical points, which are crucial for identifying important features of the graph.

Critical Points

Critical points occur where the derivative is either zero or undefined. These points are where the function might have local maxima or minima.

• For \( y = x^{2/5} \), the derivative \( \frac{2}{5}x^{-3/5} \) is undefined at \( x = 0 \).
• The derivative does not equal zero for any other \( x \), so the only critical point is at \( x = 0 \).

At this critical point, \( (0,0) \), we further analyze to see if it corresponds to a local extremum by considering the behavior of the derivative around this point. Critical points give us the chance to check for potential turning points on the graph, something very important when sketching the behavior of functions.

Extremum

An extremum refers to either a maximum or minimum point on a function's graph. In this case, we focus on local extrema.

• The function \( y = x^{2/5} \) has a local minimum at the critical point \( (0, 0) \).
• Because this function is never negative, this is also the absolute minimum on the interval where it is defined.

Extrema are significant because they represent the highest or lowest points within a certain range on the graph. They can indicate where functions attain peaks or valleys, and understanding this aids in interpreting and graphing the function.

Inflection Points

Inflection points are where the curve changes its concavity, or how it "bends." To find them, we look at the second derivative.

• The second derivative for \( y = x^{2/5} \) is \( -\frac{6}{25}x^{-8/5} \).
• This second derivative is undefined at \( x=0 \) but remains negative for \( x eq 0 \) which means no sign change occurs.

Therefore, the function lacks inflection points since it stays concave down across its domain. Understanding inflection points lets us visualize where the graph of a function might change its direction of bending; however, not all functions include these transitions in their behavior.