Problem 50 - Thomas Calculus

Question

In Exercises \(41-50,\) determine all critical points for each function. $$ y=x-3 x^{2 / 3} $$

Step-by-Step Solution

Answer

Critical points are at \( x = 8 \) and \( x = 0 \).

Step 1: Find the First Derivative

To determine critical points, we first need to find the derivative of the function to identify where the slope is zero. The given function is \( y = x - 3x^{2/3} \). We'll differentiate it with respect to \( x \). Using the power rule, the derivative of \( x \) is \( 1 \), and the derivative of \(-3x^{2/3}\) is \(-3 \cdot \frac{2}{3}x^{-1/3} = -2x^{-1/3} \).Thus, the derivative \( y' = 1 - 2x^{-1/3} \).

Step 2: Set the First Derivative to Zero

Critical points occur where the first derivative is zero or undefined. We first set the derivative to zero to find the points where the slope is zero. \[ 1 - 2x^{-1/3} = 0 \] Solve for \( x \):\[ 1 = 2x^{-1/3} \] \[ x^{-1/3} = \frac{1}{2} \] To isolate \( x \), raise both sides to the power of \(-3\):\[ x = \left( \frac{1}{2} \right)^{-3} = 8 \].

Step 3: Check where the Derivative is Undefined

Besides where the derivative is zero, critical points also occur where the derivative is undefined. The derivative \( y' = 1 - 2x^{-1/3} \) is undefined when the term \( x^{-1/3} \) is undefined, which occurs when \( x = 0 \).

Step 4: List the Critical Points

From the previous steps, we have found two critical points: \( x = 8 \) where the derivative is zero, and \( x = 0 \) where the derivative is undefined.

Key Concepts

First Derivative

When you hear about the first derivative, it's all about finding the slope of a function at any point. In simple terms, it tells us how fast or slow something is changing, like the speed of a car at a given second. Given the function in our exercise, which is \( y = x - 3x^{2/3} \), finding the first derivative means calculating how the function behaves as \( x \) changes.
To find the first derivative, we apply rules of differentiation. Here, we used the power rule. For \( x \), the derivative is 1 because \( d(x)/dx = 1 \). For \( -3x^{2/3} \), it requires a bit more effort:

• Bring down the exponent: \(-3 \cdot \frac{2}{3}\)
• Subtract one from the exponent: resulting in \(-x^{-1/3}\)

Putting it all together, the first derivative is \( y' = 1 - 2x^{-1/3} \). This tells us how the function changes and sets the stage for finding critical points.

Derivative of a Function

A derivative of a function gives you a new function that shows the rate of change of the original function. It's a central concept in calculus that extends our understanding of motion, growth, and decline.
With derivatives, we can:

• Calculate how things change over time or in space, like the velocity of a moving object.
• Locate critical points in functions where significant changes occur, such as peaks, troughs, or other turning points.

The derivative helps unlock the behavior of any function with respect to its variable(s). For our given function, \( y = x - 3x^{2/3} \), its derivative was \( y' = 1 - 2x^{-1/3} \). While this function's change might look simple at first glance, its derivative exposes how it speeds up or slows down.
By analyzing the first derivative, we determine areas of rapid change or where the function remains constant. That's why finding places where the derivative equals zero or is undefined is crucial. It shows where the function has critical behavioral shifts.

Undefined Derivative

Sometimes, taking the derivative of a function leads us to a scenario where the derivative is undefined. This specifically happens when you encounter terms that don't have real solutions or introduce terms that break the rules, like dividing by zero.
In the context of our function, \( y = x - 3x^{2/3} \), the derivative \( y' = 1 - 2x^{-1/3} \) becomes undefined at certain points. Here, the term \( x^{-1/3} \) is undefined whenever \( x = 0 \). That's because raising zero to a negative power results in division by zero, which is undefined in mathematics.
Finding where the derivative is undefined is just as important as finding where it is zero. These undefined spots can indicate important features or breaks in the function's graph.
For our task, we discovered critical points at \( x = 8 \) where the slope equals zero and at \( x = 0 \) where the derivative is undefined. These result in sharp changes or discontinuities within the function.