Problem 33 - Single Variable Calculus

Question

In Exercises \(25-40\), find the critical number \((s)\), if any, of the function. $$ f(x)=x^{2 / 3} $$

Step-by-Step Solution

Answer

The only critical number of the function \(f(x) = x^{\frac{2}{3}}\) is \(x = 0\).

Step 1: Find the derivative of the function

We are given the function $$f(x) = x^{\frac{2}{3}}$$. To find the derivative, we will use the power rule of differentiation: \[ \frac{d}{dx}(x^n) = nx^{n-1} \] So, $$f'(x) = \frac{2}{3} \cdot x^{\frac{2}{3} - 1} = \frac{2}{3} \cdot x^{-\frac{1}{3}}$$

Step 2: Determine critical numbers where the derivative is zero.

Now, we will try to find the values of x for which $$f'(x) = 0$$. However, notice that the function $$f'(x)$$ has a negative power. As such, it can never be equal to 0.

Step 3: Determine critical numbers where the derivative does not exist.

Since $$f'(x) = \frac{2}{3} \cdot x^{-\frac{1}{3}}$$, the derivative does not exist when $$x^{-\frac{1}{3}}$$ is undefined. This happens when x is equal to 0. So, x = 0 is a critical number of the function. To summarize, the only critical number of the function $$f(x) = x^{\frac{2}{3}}$$ is $$x = 0$$.

Key Concepts

Power Rule of Differentiation

Understanding the power rule of differentiation is essential for anyone studying calculus. It is a simple yet powerful tool that allows you to take the derivative of a function that is a power of x. The rule states that if you have a function of the form f(x) = x^n, where n is any real number, the derivative of this function is f'(x) = nx^(n-1). This means you multiply the original exponent by the coefficient and then subtract one from the exponent. When applied to f(x) = x^(2/3), as in our exercise, we use the power rule to find f'(x) = (2/3)*x^(2/3 - 1) = (2/3)*x^(-1/3).

This technique simplifies differentiation and is frequently the first step in finding critical points, which are points where the function's slope is zero or the derivative is undefined. It's important to fully grasp this concept as it makes calculus more manageable and provides the foundation for many of the problems you'll encounter.

Derivative of a Function

The derivative of a function at a point measures the rate at which the function's value changes at that point. It's a fundamental concept in calculus and essential for understanding the behaviour of functions. To calculate a derivative, we apply rules like the power rule, product rule, and chain rule, depending on the form of the function. In the context of our example, f(x) = x^(2/3), we are interested in how rapidly the value of the function changes with respect to x. The derivative we obtained, f'(x) = (2/3)*x^(-1/3), tells us the function's slope at any given point except where it is undefined. The derivative allows us to find slopes, rates, and many other important concepts that describe the dynamic nature of things. An intuitive understanding of derivatives forms the core of calculus and its applications.

Finding Critical Points

Critical points are vital in studying a function's behavior since they are where the function's derivative is either zero or undefined, often corresponding to peaks, troughs, or inflection points on a graph. To find the critical points of a function, you typically set the derivative equal to zero and solve for x, as well as look for where the derivative is undefined.

In our exercise example, we have the derivative f'(x) = (2/3)*x^(-1/3). Setting this equal to zero is unproductive, as a fraction with a non-zero numerator over an exponent of x cannot equal zero. Therefore, we must look for where the derivative is undefined, which occurs at x = 0 because we cannot divide by zero. Consequently, x = 0 is the critical point of the function f(x) = x^(2/3). It is imperative to understand both the algebraic methods and the conceptual reasons behind finding critical points to thoroughly analyze a function's behavior.