Question

Follow the instructions for Problem 2 using \(f(x)=x(1-x)\) for \(0

Short answer

To summarize, the given function \(f(x) = x(1-x)\) within the domain \((0,1)\) has the critical point at \(x=\frac{1}{2}\). The function is increasing in the interval \((0,\frac{1}{2})\) and decreasing in the interval \((\frac{1}{2},1)\). The maximum value of the function is \(\frac{1}{4}\) and the minimum value is \(0\). Therefore, the range of the function is \((0, \frac{1}{4}]\).

Step-by-step solution

Identify the Domain of the function

The domain of the function \(f(x) = x(1-x)\) is given as \(0<x<1\). This means that \(x\) can take any value between 0 and 1, excluding 0 and 1. So the domain is: \((0,1)\)

Determine the Range of the function

To determine the range, we first need to find the maximum and minimum values that this function can attain. We'll do this by finding its critical points and analyzing the function within the given domain.

Find the critical points of the function

To find the critical points, we need to find the derivative of the function \(f(x)\), set it equal to 0, and solve for \(x\). This will give us the points of the function where it changes from increasing to decreasing or vice versa. The derivative of \(f(x) = x(1-x)\) is given by: $$f'(x) = (1-x) + (-x) = 1 - 2x$$ Now, set \(f'(x)=0\) and solve for \(x\): $$1-2x=0 \implies x=\frac{1}{2}$$ So, our critical point is \(x=\frac{1}{2}\).

Analyze the function for increasing and decreasing intervals

Now that we have the critical point, it's time to check whether the function is increasing or decreasing within those intervals. We can do this by plugging values of \(x\) from different intervals into the derivative. - Choose a test point from the interval \((0,\frac{1}{2})\), say \(x=0.25\). Plug it into the derivative, \(f'(0.25)=1-2(0.25)=0.5>0\). This means that the function is increasing in the interval \((0,\frac{1}{2})\). - Choose a test point from the interval \((\frac{1}{2},1)\), say \(x=0.75\). Plug it into the derivative, \(f'(0.75)=1-2(0.75)=-0.5<0\). This means that the function is decreasing in the interval \((\frac{1}{2},1)\).

Determine the maximum and minimum values of the function

Now that we know that the function is increasing in the interval \((0,\frac{1}{2})\) and decreasing in the interval \((\frac{1}{2},1)\), we can conclude that the maximum value of the function is attained at the critical point. To find the maximum value, plug the critical point value into the function: \(f(\frac{1}{2}) = \frac{1}{2}(1-\frac{1}{2}) = \frac{1}{4}\). As the function is defined only within the domain \((0,1)\) and it is always positive within this domain, the minimum value attained by the function is at either end. Since \(0\) and \(1\) are not included in the domain, the minimum value will be \(0\). Therefore, the range of the function is \((0, \frac{1}{4}]\).

Key concepts

Domain of a Function

The domain of a function represents all the possible input values (generally represented as 'x') for which the function is defined. It's like the 'allowed' field where the function can play without stumbling into undefined behaviors, such as division by zero or taking the square root of a negative number.

For the given example where the function is defined as \( f(x) = x(1-x) \) with the condition that \( 0 < x < 1 \), the domain is an open interval represented as \( (0, 1) \). This implies we are looking at all the numbers between 0 and 1, not including the endpoints themselves. Imagine a number line where you highlight every point between 0 and 1, but the points at 0 and 1 remain unmarked; that is the visual representation of our function's domain.

Range of a Function

Conversely, the range of a function is the set of all output values (often represented as 'y' or \( f(x) \)) that can be produced given the domain. Figuratively speaking, it is the extent of vertical reach your function has on a graph.

For \( f(x) = x(1-x) \), we observed that the function's maximum occurs at a critical point. In Step 5 of the solution, it was determined that the function has a maximum value of \( \frac{1}{4} \) which occurs at the critical point \( x=\frac{1}{2} \). Since the function is positive and continuous on the open interval \( (0,1) \) and approaches zero as it nears the endpoints, the range is the closed interval \( (0, \frac{1}{4}] \), with the smallest value being slightly greater than zero (as the endpoints are not included) and the largest being \( \frac{1}{4} \) itself.

Critical Points

Critical points are like the strategic high grounds in the landscape of a function. They are the x-values where the function's derivative is zero or undefined, indicating potential peaks, troughs, or plateaus in the graph of the function.

In our function, the critical point arises by setting the derivative \( f'(x) = 1 - 2x \) equal to zero. Solving for \( x \), we got the critical point at \( x = \frac{1}{2} \). This is where our function switches from climbing to descending; thus, it marks the highest point in this domain, much like a hilltop within the boundaries of a map.

Function Analysis

Analyzing a function involves looking at its domain, range, critical points, and understanding the behavior such as intervals of increase or decrease, concavity, and points of inflection. It's like conducting a full health check-up of the function, ensuring we know how it behaves at every point.

In our case, we found that \( f(x) = x(1-x) \) was increasing on the interval \( (0, \frac{1}{2}) \) and decreasing on \( (\frac{1}{2}, 1) \). This analysis is crucial as it not only tells us where the function reaches its maxima and minima but also how it gets there. The fact that our function increases to the critical point and then decreases means it had one peak within our domain, and that's essential information for sketching graphs or predicting function values.