Problem 46 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test or the Second Derivative Test to determine the relative extreme values, if any, of the function. Then sketch the graph of the function. $$ f(x)=\sqrt{x-x^{2}} $$

Step-by-Step Solution

Answer

The function \( f(x) \) has a relative maximum at \( x = 0.5 \).

Step 1: Find the Domain of the Function

The function is defined as \( f(x) = \sqrt{x-x^2} \). The expression inside the square root, denoted by \( g(x) = x - x^2 \), must be non-negative for \( f(x) \) to be real-valued. Solving \( x - x^2 \geq 0 \), we factor as \( x(1-x) \geq 0 \). Therefore, the solutions are \( 0 \leq x \leq 1 \). Thus, the domain of \( f(x) \) is \( [0, 1] \).

Step 2: First Derivative and Critical Points

The first derivative \( f'(x) \) is needed to find critical points. We use the chain and product rules: \( f'(x) = \frac{d}{dx}(x-x^2)^{1/2} = \frac{1}{2}(x-x^2)^{-1/2}(1-2x) \). This simplifies to \( f'(x) = \frac{1-2x}{2\sqrt{x-x^2}} \). We set \( f'(x) = 0 \) to find critical points, resulting in \( 1-2x = 0 \), thus, \( x = 0.5 \).

Step 3: Second Derivative and Concavity

Now determine whether the critical point is a maximum or minimum using the second derivative test. First, calculate the second derivative \( f''(x) \). For simplicity using results from \( f'(x) = \frac{1-2x}{2\sqrt{x-x^2}} \), we compute the second derivative \( f''(x) \) and find that \( f''(0.5) < 0 \). This indicates that \( x=0.5 \) is a relative maximum point.

Step 4: Evaluate Function at Critical Points and Endpoints

Evaluate \( f(x)\) at the critical point and endpoints, \( x = 0, 0.5, \, \text{and} \, 1 \). Calculate \( f(0) = \sqrt{0-0^2} = 0 \),\( f(0.5) = \sqrt{0.5-0.5^2} = \sqrt{0.25} = 0.5 \), and \( f(1) = \sqrt{1-1^2} = 0 \). The relative maximum of the function is 0.5 at \( x=0.5 \), with no further local minima or maxima.

Step 5: Sketch the Graph

Based on the evaluations, the function starts at point (0,0), peaks at (0.5,0.5), and returns to (1,0). Thus, the sketch is a concave down arc from 0 to 1 with a maximum at (0.5, 0.5).

Key Concepts

Second Derivative Test

The Second Derivative Test is a useful method to determine the concavity of a function and classify critical points as relative minima or maxima. Once you have found a critical point, you can apply the test by evaluating the second derivative at this point. Here’s how it works:

• If \( f''(x) > 0 \) at a critical point, the function is concave up, and the point is a relative minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a relative maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and you may need to use another test or consider higher order derivatives.

In our solution, we found that \( f''(0.5) < 0 \), which tells us the function at \( x = 0.5 \) is concave down. Therefore, there is a relative maximum at this point.

relative extreme values

Relative extreme values refer to the highest or lowest values that a function takes on within a certain interval or neighborhood. These values are of special interest because they inform us about the local behavior of a function.To identify relative extrema, we typically follow these steps:

• First, find the critical points of the function within its domain. Critical points occur where the first derivative is zero or undefined.
• Second, apply a derivative test, like the First Derivative or the Second Derivative Test, to classify these critical points as relative maxima, minima, or saddle points.

In the provided exercise, the function \( f(x) = \sqrt{x-x^2} \) has a relative maximum at \( x = 0.5 \) where the value of the function is 0.5. The evaluations at the endpoints reveal that there are no further relative extrema within the interval \([0, 1]\).

critical points

Critical points are values of \( x \) in the domain of a function where the first derivative \( f'(x) \) is zero or undefined. These points are essential in finding potential maxima and minima because they indicate spots where the function's rate of change is momentarily flat or nonexistent.To identify critical points:

• Calculate the first derivative \( f'(x) \).
• Solve \( f'(x) = 0 \) and check for points where the derivative does not exist, within the function’s domain.

In our case, plugging \( f'(x) = \frac{1-2x}{2\sqrt{x-x^2}} \) into zero gave us \( x = 0.5 \), the critical point over the interval \([0, 1]\). This has been verified through further calculations and checking against the second derivative.

function domain

The domain of a function defines the set of input values (\( x \)) for which the function is defined. Understanding this is foundational before performing further analysis like finding derivatives or extrema, as operations must stick to this range.For functions involving square roots, such as \( f(x) = \sqrt{x-x^2} \), it’s crucial that expressions under the root are non-negative since square roots of negative numbers are not real. To find the domain:

• Solve \( x - x^2 \geq 0 \).
• This factors to \( x(1-x) \geq 0 \).
• Evaluate the solutions to determine \( x \) ranges, resulting in \( 0 \leq x \leq 1 \).

Thus, the domain of our function is \([0, 1]\), reflecting all possible values of \( x \) where the function is valid and defined.