Problem 48 - Thomas Calculus in SI Units

Question

Determine all critical points for each function. $$g(x)=\sqrt{2 x-x^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Find the derivative

To find the critical points, we start by finding the derivative of the function using the chain rule. The function is given by \[ g(x) = \sqrt{2x - x^2} \]. We can write this as \[ g(x) = (2x - x^2)^{1/2} \]. The derivative is given by: \[ g'(x) = \frac{1}{2}(2x - x^2)^{-1/2} \cdot (2 - 2x) \].Simplifying, we have: \[ g'(x) = \frac{2 - 2x}{2\sqrt{2x - x^2}} \],which further simplifies to \[ g'(x) = \frac{1 - x}{\sqrt{2x - x^2}} \].

Step 2: Set the derivative to zero

Next, we set the derivative to zero to find critical points:\[ \frac{1 - x}{\sqrt{2x - x^2}} = 0 \].The fraction is zero when the numerator equals zero. So, solve \[ 1 - x = 0 \].Thus, \[ x = 1 \].

Step 3: Check the domain of the function

The function \[ g(x) = \sqrt{2x - x^2} \] requires that the expression under the square root be non-negative:\[ 2x - x^2 \geq 0 \]. Factoring gives:\[ x(2 - x) \geq 0 \]. The solutions are \[ x \leq 0 \] or \[ x \geq 2 \].However, testing some points shows that the valid domain is \[ 0 \leq x \leq 2 \].

Step 4: Verify the critical point is within the domain

We found that \[ x = 1 \] is a critical point. It is important to check that this point is within the domain of the function. Since \[ 0 \leq 1 \leq 2 \],\[ x = 1 \] is indeed within the domain of \[ g(x) \].

Key Concepts

Chain Rule

When we find the derivative of complex functions, like the square root function in this exercise, we often apply the chain rule, which is an essential calculus tool. The chain rule is used when a function is composed of two or more functions.
For instance, in our example, we have \[ g(x) = (2x - x^2)^{1/2} \], where the inner function is \( 2x - x^2 \) and the outer function is \( u^{1/2} \) (imagine \( u = 2x-x^2 \)).

• First, take the derivative of the outer function with respect to the inner: \[ \frac{1}{2}u^{-1/2} \].
• Second, multiply by the derivative of the inner function, \( 2x - x^2 \), which is \( 2 - 2x \).
• The result of this process is the derivative, \[ g'(x) = \frac{1}{2}(2x - x^2)^{-1/2} \cdot (2 - 2x) \].

This method allows us to differentiate effectively by breaking complicated structures into simpler parts.

Derivative

A derivative represents the rate at which a function changes. It is a fundamental concept in calculus. In practical terms, it tells us the slope of the tangent line at any point on a graph of the function.
For instance, finding \( g'(x) \) for \( g(x) = \sqrt{2x - x^2} \) means we need the slope of tangents to this curve, revealing how fast \( g(x) \) is changing at each \( x \) value.### Steps to Calculate the Derivative1. **Rewriting for Easier Differentiation**: Begin by rewriting square root functions as a power. For example, \( \sqrt{2x - x^2} \) as \( (2x - x^2)^{1/2} \).2. **Apply Calculus Rules**: Use differentiation techniques like the power rule and chain rule.3. **Simplify**: Break down mathematical expressions further for ease of use, resulting in \[ g'(x) = \frac{1 - x}{\sqrt{2x - x^2}} \].
Determining derivatives gives insights into the function's increasing or decreasing behavior.

Function Domain

The domain of a function consists of all possible input values \( x \) for which the function \( g(x) \) is defined. It is crucial to determine where a function is valid, especially for expressions involving square roots.
For \( g(x) = \sqrt{2x-x^2} \), the expression under the root must be non-negative to ensure real values. This leads us to check:

• \[ 2x - x^2 \geq 0 \]
• Factorize to \[ x(2 - x) \geq 0 \].
• Solve for the boundary values, discovering \[ x \leq 0 \] or \[ x \geq 2 \].

Further testing shows the valid domain is confined to \( 0 \leq x \leq 2 \). Always confirming domain helps avoid undefined scenarios and leads to proper point evaluation.

Non-negative Constraint

In the context of square roots, which are inherently real and non-negative, imposing a non-negative constraint ensures validity. For functions like \( g(x) = \sqrt{2x-x^2} \), this requirement leads to \( 2x - x^2 \geq 0 \).
This inequality ensures that the square root function does not encounter the impossible square root of a negative number, which is undefined in real numbers. Let's break it down further:

• Factor out the expression: \[ x(2 - x) \].
• Determine where this product is zero or positive. Critical points arise where each factor is zero: \( x = 0 \) and \( x = 2 \).
• Evaluate intervals between critical points to consider changes in sign:
• Examine test points to confirm valid regions, resulting in \( 0 \leq x \leq 2 \) as valid.

This approach guarantees correct \( x \) values for \( g(x) \) to have real results.