Problem 45 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The critical point is \(x = 1\).

Step 1: Differentiate the Function

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

Step 2: Find the Critical Points

Critical points occur where the derivative is zero or undefined. We need to solve \[ 2x - \frac{2}{x^2} = 0 \] This simplifies to: \[ 2x = \frac{2}{x^2} \] Multiply both sides by \(x^2\) to eliminate the fraction: \[ 2x^3 = 2 \] Divide both sides by 2: \[ x^3 = 1 \] Take the cube root of both sides to find \(x\): \[ x = 1 \] The expression \(2x - \frac{2}{x^2}\) is also undefined when \(x = 0\), but \(x = 0\) is not in the domain of the original function, so it is not a critical point.

Step 3: Verify the Critical Point

After finding the critical point \(x = 1\), verify it is within the domain of the original function. The function \(y = x^2 + \frac{2}{x}\) is undefined at \(x = 0\), but \(x = 1\) is within the domain \(x eq 0\). Thus, \(x = 1\) is a valid critical point for this function.

Key Concepts

Derivative

In mathematics, the term 'derivative' refers to the measure of how a function changes as its input changes. It serves as a fundamental tool for understanding the behavior of functions and is typically denoted as \(y'\) or \frac{dy}{dx}\.To find a derivative, we examine the rate of change of a function. For example, if we have a function \(y = x^2 + rac{2}{x}\), its derivative, represented as \(y'\), will tell us how \(y\) changes with respect to \(x\). For this function, the steps would be:

• Find the derivative of \(x^2\), which is \(2x\).
• Find the derivative of \(rac{2}{x}\), which, using the rules of differentiation, is \(-rac{2}{x^2}\).

Combining these results, the derivative of the function becomes \(y' = 2x - rac{2}{x^2}\). This expression is crucial as it helps us identify critical points and analyze the function's behavior.

Differentiation

Differentiation is the process of finding the derivative of a function. It involves applying specific rules to obtain the derivative, which represents how a function's output value changes concerning its input value. In our example of the function \(y = x^2 + rac{2}{x}\), differentiation allows us to find out how the function behaves by calculating its derivative.Here are the steps involved in differentiating the given function:

• Firstly, apply the differentiation rules to each term of the function.
• Compute separately the derivative of each component, such as \(x^2\) and \(rac{2}{x}\).
• Lastly, combine these derivatives by algebraic addition or subtraction to get the complete derivative expression.

Thus, through differentiation, we derive \(y' = 2x - rac{2}{x^2}\), which can be used for further analysis of functions like locating and understanding critical points.

Power Rule

The power rule is one of the most common rules of differentiation you will often encounter. It simplifies the process of finding the derivative of polynomial functions. The rule states that if you have a function of the form \(f(x) = x^n\), its derivative is \(rac{d}{dx} x^n = nx^{n-1}\).This rule makes it straightforward to differentiate terms like \(x^2\):

• Identify the power \(n\) in the term \(x^2\), which is 2.
• Multiply the term by the power: \(2 imes x^{2-1} = 2x\).

For our example function, we used the power rule to find the derivative of \(x^2\) quickly, resulting in \(2x\). It's essential to apply the power rule cautiously, especially when dealing with negative or fractional exponents, as it applies universally across all polynomial types.