Problem 40 - Algebra and Trigonometry

Question

\(35-42\) A function is given. (a) Find all the local maximum and minimum values of the function and the value of \(x\) at which each occurs. State each answer correct to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer correct to two decimal places. $$ U(x)=x \sqrt{x-x^{2}} $$

Step-by-Step Solution

Answer

Local max at \(x = 0.75\). Increase on \((0, 0.75)\), decrease on \((0.75, 1)\).

Step 1: Find the Derivative

First, we find the derivative of the function \( U(x) = x \sqrt{x-x^2} \). To differentiate this, we can use the product rule and the chain rule.Rewrite the function as \( U(x) = x (x - x^2)^{1/2} \). Differentiate:\[U'(x) = \frac{d}{dx} \left( x \right) (x-x^2)^{1/2} + x \cdot \frac{d}{dx} (x-x^2)^{1/2}\]Thus:\[U'(x) = (x-x^2)^{1/2} + x \cdot \frac{1}{2} (x-x^2)^{-1/2} \cdot (1 - 2x)\]

Step 2: Simplify and Set Derivative to Zero

Simplify the derivative:\[U'(x) = (x-x^2)^{1/2} + \frac{x (1-2x)}{2 \sqrt{x-x^2}}\]For critical points, set \( U'(x) = 0 \). We multiply through by \( 2 \sqrt{x-x^2} \) to clear the fraction:\[2(x-x^2) + x(1-2x) = 0\]\[2x - 2x^2 + x - 2x^2 = 0\]\[-4x^2 + 3x = 0\]Factor:\[x(3 - 4x) = 0\]

Step 3: Find Critical Points

Solve the factorized equation:\[x = 0 \quad \text{or} \quad 3 - 4x = 0\]Solving \(3 - 4x = 0\) gives:\[x = \frac{3}{4}\]Thus, the critical points are \(x = 0\) and \(x = 0.75\).

Step 4: Determine Second Derivative for Concavity

To determine whether these points are maximum or minimum, find the second derivative \( U''(x) \). Take the derivative of \( U'(x) \):\[U''(x) = d/dx \left((x-x^2)^{1/2} + \frac{x (1-2x)}{2 \sqrt{x-x^2}} \right)\]This is complicated to compute exactly without computational tools, but we can test intervals around the critical points to determine concavity.

Step 5: Interval Test for Maximum or Minimum

Using a test interval approach around critical points:- Test \( x = 0.25 \) - net positive value, indicates increasing behavior moving away from 0.- Test \( x = 0.5 \) and \( x =1\) - confirms switch from increasing to decreasing from 0.75.From this logic, \( x = 0.75 \) is a local maximum, and \( x = 0 \) is a neither.

Step 6: Determine Intervals of Increase or Decrease

Using critical points, determine intervals:- Increasing on \((0, 0.75)\) as behavior goes from positive, and tone sets to negative.- Decreasing on \((0.75, 1)\).

Step 7: State Final Answers

Local maximum at \( x = 0.75 \) with value \( U(0.75) = 0.433 \). No local minimum.Increasing on interval \((0, 0.75)\), decreasing on interval \((0.75, 1)\).

Key Concepts

Derivative

Derivatives are a fundamental tool in calculus that represent the rate of change or the slope of a function at any given point. In simple terms, if you imagine a curve plotted on a graph, the derivative at a particular point is the steepness or incline of the tangent at that point. Derivatives help us to understand how a function behaves, showing whether it's increasing, decreasing, or leveling out.

To find the derivative of a function like \[ U(x) = x \sqrt{x-x^2} \]one can use rules such as the product rule, which is applicable when the function is a product of two simpler functions. The chain rule also comes into play when dealing with compositions of functions.

- The **Product Rule** is: \( (uv)' = u'v + uv' \).- The **Chain Rule** gives the derivative of a composite function: \( (f(g(x)))' = f'(g(x)) \, g'(x) \). In this exercise, we apply these to get the derivative \[ U'(x) = (x-x^2)^{1/2} + \frac{x (1-2x)}{2 \sqrt{x-x^2}} \].Finding where this derivative equals zero helps to identify critical points, which could be local maxima or minima.

Local Maximum and Minimum

Local maxima and minima are points on a graph where a function reaches its highest or lowest value locally, meaning within a small surrounding interval. These points are crucial in analyzing the behavior of a function, as they help determine where a function curves upwards or downwards.

To identify these points, we set the derivative to zero, \(U'(x) = 0\), and solve for \(x\). In this example, the critical points were found to be \(x = 0\) and \(x = 0.75\). However, we need to differentiate further by understanding whether these points are maxima or minima by using the second derivative test.

- If \(U''(x) > 0\) at a critical point, the point is a local minimum.- If \(U''(x) < 0\), it indicates a local maximum.Examining the function's behavior around the critical points, it was observed that at \(x = 0.75\), there is a transition from increasing to decreasing, signifying a local maximum. There's no local minimum here since the conditions don't match.

Increasing and Decreasing Intervals

An increasing interval on a graph is where the function's values go up as \(x\) increases, and a decreasing interval is where the values go down as \(x\) increases. Determining these intervals reveals where the function rises or falls.

For this, we use the sign of the derivative:- **Positive derivative (\(U'(x) > 0\)):** The function is increasing.- **Negative derivative (\(U'(x) < 0\)):** The function is decreasing.After finding the critical points, test intervals between and around these points.

- Between \(x = 0\) and \(x = 0.75\), the derivative is positive, indicating the function is increasing.- After \(x = 0.75\) to \(x = 1\), the derivative turns negative, showing a decreasing trend.Thus, the function \(U(x)\) increases within the interval \((0, 0.75)\) and decreases within \((0.75, 1)\). These observations define the overall behavior of the function across specified intervals.