Problem 27 - Calculus A New Horizon

Question

Use any method to find the relative extrema of the function \(f\). $$f(x)=x^{4 / 5}$$

Step-by-Step Solution

Answer

The function has no relative extrema but a point of inflection at \( x = 0 \).

Step 1: Determine the Derivative

To find the relative extrema, we first need to find the derivative of the function. Given the function \[ f(x) = x^{4/5} \]we use the power rule to differentiate:\[ f'(x) = \frac{4}{5} x^{(-1/5)} \].

Step 2: Set the Derivative to Zero

We set the derivative equal to zero to find the critical points:\[ \frac{4}{5} x^{(-1/5)} = 0 \].Since a non-zero constant multiplied by a variable cannot make the equation zero, there are no solutions from this equation. However, it's important to check the point where the derivative might be undefined.

Step 3: Identify Points of Interest

Since the derivative \[ f'(x) = \frac{4}{5} x^{-1/5} \] is undefined when \( x = 0 \), we consider \( x = 0 \) as a critical point to check for relative extrema.

Step 4: Evaluate Extrema Using the Second Derivative Test

To use the second derivative test, we first find the second derivative:\[ f''(x) = \frac{4}{5} \cdot \frac{-1}{5} x^{-6/5} = -\frac{4}{25} x^{-6/5} \].Evaluate at the critical point \( x = 0 \):Since \( f''(x) \) is undefined at \( x = 0 \) and negative for \( x > 0 \), and positive for \( x < 0 \), we analyze the behavior around \( x = 0 \).- For \( x > 0 \), \( f'(x) \) indicates a decreasing slope.- For \( x < 0 \), \( f'(x) \) indicates an increasing slope.Hence, \( x = 0 \) is not a relative extrema but a point of inflection.

Key Concepts

Relative Extrema

Relative extrema are points on a graph where a function reaches either a minimum or maximum value within a certain interval. Unlike absolute extrema, which are the highest or lowest points over the entire function, relative extrema might just be locally highest or lowest within nearby points. In this context, it's important to understand that a relative maximum is where the function shifts from increasing to decreasing, while a relative minimum is where it shifts from decreasing to increasing. To find these critical points, we typically look for where the first derivative is zero or undefined.

Derivatives

Derivatives are a fundamental concept in calculus that represent the rate of change of a function. If you think of a graph, the derivative at any given point can be visualized as the slope of the tangent line to the curve at that point. For example, if you are driving a car, the speedometer is showing you the derivative of your position with respect to time. In mathematical terms, the derivative helps us find how quickly or slowly a function is changing at any given value of the independent variable. This change is crucial when identifying relative extrema, as it helps indicate potential peaks or troughs in the graph.

Critical Points

Critical points are specific values in the domain of a function where its derivative is either zero or undefined. These points are significant as potential locations for relative extrema. In the process of finding critical points, you first take the derivative of your function and then solve where this derivative equals zero. However, sometimes the derivative might not exist at certain points, as was the case with the function examined above, where it was undefined at point \( x = 0 \). Such points must also be checked for extrema by analyzing the behavior of the function around them.

Power Rule

The power rule is a quick and essential technique for finding derivatives of functions involving powers of \( x \). It states that if you have a function \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \). This rule simplifies the process of differentiation significantly. For example, the derivative of \( x^{4/5} \) turns out to be \( \frac{4}{5}x^{-1/5} \) using the power rule, which then facilitates finding critical points by setting this derivative either to zero or determining where it may be undefined. This rule is powerful because it applies to any rational exponent, making it versatile for a range of differentiable functions.