Problem 45 - Thomas Calculus in SI Units

Question

Determine all critical points for each function. $$y=x^{2}+\frac{2}{x}$$

Step-by-Step Solution

Answer

The critical points are at \( x = 1 \) and \( x = 0 \).

Step 1: Find the First Derivative

To find the critical points, first we need to find the derivative of the function. The given function is:\[ y = x^2 + \frac{2}{x} \]The derivative, using the power rule and the derivative of \( \frac{1}{x} \), is:\[ y' = 2x - \frac{2}{x^2} \]

Step 2: Set the First Derivative to Zero

Critical points occur where the derivative is zero or undefined. Set the derivative to zero to find the critical points:\[ 2x - \frac{2}{x^2} = 0 \]

Step 3: Solve the Equation

Solve the equation from Step 2:First, clear the fraction by multiplying both sides by \( x^2 \):\[ 2x \cdot x^2 - 2 = 0 \]\[ 2x^3 - 2 = 0 \]Then, solve for \( x \):\[ 2x^3 = 2 \]\[ x^3 = 1 \]\[ x = 1 \]

Step 4: Check for Undefined Derivative

Find where the derivative is undefined, which occurs if the denominator is zero. Here, the denominator is \( x^2 \), which is zero at \( x = 0 \). The derivative is undefined at \( x = 0 \), which is another critical point.

Step 5: List the Critical Points

The critical points are the values of \( x \) where the derivative is zero or undefined. From our calculations, these are:- \( x = 1 \) (where the derivative is zero)- \( x = 0 \) (where the derivative is undefined)

Key Concepts

First Derivative

The first derivative of a function provides us with the necessary information to determine the critical points which are essential in understanding the function's behavior. In this exercise, we start with the function: \[ y = x^2 + \frac{2}{x} \]To find the first derivative, we apply differentiation rules. The first derivative of a function, denoted as \( y' \) or \( \frac{dy}{dx} \), tells us the rate of change of the function at any point on its curve.
Finding the first derivative is the first step toward identifying critical points, as these are values of \( x \) where the function could turn around or have a slope of zero.

• Calculate the derivative using rules such as the power rule and the derivative of reciprocal functions.
• This provides the expression \( y' = 2x - \frac{2}{x^2} \).

Power Rule

The power rule is a simple and powerful tool for finding the derivative of a polynomial. For any function \( f(x) = x^n \), the derivative is found by multiplying the exponent \( n \) by the coefficient and subtracting one from the exponent: \[ f'(x) = nx^{n-1} \]This rule allows for quick calculations, making it easier to handle various polynomial terms within a function.
In our exercise:

• For the term \( x^2 \), the derivative using the power rule is \( 2x \).
• For the term \( \frac{2}{x} \), we first rewrite it as \( 2x^{-1} \) to apply the power rule.
• Using the power rule, the derivative becomes \(-2x^{-2} \), which is \(-\frac{2}{x^2} \).

Rewriting fractions with negative exponents helps in applying the power rule, simplifying the expression, and ultimately finding the derivative.

Derivative Undefined

A derivative can be undefined at certain points, which is just as important as where it equals zero when finding critical points. In our problem, the expression for the derivative is:\[ y' = 2x - \frac{2}{x^2} \]The denominator \( x^2 \) creates a scenario where the derivative cannot be calculated if it equals zero.
Points where a derivative is undefined often indicate vertical tangent lines or potential asymptotic behavior.

• For the derivative, \( y' \), it is undefined at \( x = 0 \), because division by zero is impossible.
• By identifying these undefined points, we can locate critical points that highlight significant behavior changes in the function's graph.

Understanding both where the derivative is zero and where it is undefined equips us to fully analyze the behavior of functions and determine all critical points.