Problem 38 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the critical numbers and the open intervals on which the function is increasing or decreasing. (Hint: Check for discontinuities.) Sketch the graph of the function. $$ y=\left\\{\begin{array}{ll} -x^{3}+1, & x \leq 0 \\ -x^{2}+2 x, & x>0 \end{array}\right. $$

Step-by-Step Solution

Answer

The critical points of the function are \(x = 0\) and \(x = 1\). The function increases for \(x < 0\) and \(0 < x < 1\), and decreases after \(x > 1\). The graph of the function includes an upward-opening cubic function for \(x \leq 0\) and an downward-opening parabola for \(x > 0\).

Step 1: Analysis and Derivation

Firstly split the function into two parts as per the given conditions. Then calculate the derivative of both parts of the function. For \(x \leq 0\) we have the function \(f(x) = -x^{3}+1\) and its derivative is \(f'(x) = -3x^{2}\). For \(x > 0\) we have the function \(g(x) = -x^{2}+2x\) and its derivative is \(g'(x) = -2x + 2\).

Step 2: Determine Critical Numbers

Set the derivatives equal to 0 and find the x-values. \(-3x^2 = 0\) gives \(x = 0\) as the critical point. And \(-2x + 2 = 0\) yields \(x = 1\) as the critical point. Therefore the critical points/numbers are \(x = 0\) and \(x = 1\).

Step 3: Determine the Behavior

To find the increasing or decreasing nature, plug the values both less than and greater than the critical points into the functional form and evaluate. For \(x < 0\), we plug a negative number into \(f'(x)\) and get a positive output indicating the function is increasing. For \(0 < x < 1\), we plug a number such as 0.5 into \(g'(x)\) and get a positive output indicating the function is increasing. Lastly, for \(x > 1\), we plug a number more than 1 into \(g'(x)\) and get a negative result implying the function is decreasing after \(x = 1\).

Step 4: Sketch The Graph

For sketching the graph, plot the points and draw the shapes as per the direction of the function (increasing/decreasing). Do it separately for both parts of the function: for part \(f(x)\), it's a cube function which increases; for part \(g(x)\), it's a parabola opening downwards showing an increase at first and then a decrease.

Key Concepts

Critical Numbers

Critical numbers are essential in calculus as they help identify where a function may change its behavior, such as from increasing to decreasing. These numbers arise from setting the derivative of a function to zero or identifying where the derivative does not exist.
For the given piecewise function, we find critical numbers by solving the derivative equations of each segmented part.

• For the segment where \(x \leq 0\), the function is \(f(x) = -x^3 + 1\). The derivative is \(f'(x) = -3x^2\). Setting this equal to zero, \(-3x^2 = 0\), yields \(x = 0\).
• For the segment where \(x > 0\), the function is \(g(x) = -x^2 + 2x\). The derivative is \(g'(x) = -2x + 2\). Setting this to zero, \(-2x + 2 = 0\), gives \(x = 1\).

Together, these evaluations give us critical numbers at \(x = 0\) and \(x = 1\). These points are vital for determining the intervals of increase and decrease in the function's behavior.

Derivative

Understanding the derivative is crucial as it enables us to analyze the rate and direction of change in a function. The derivative essentially indicates how a function behaves: whether it's increasing, decreasing, or at a critical point. In mathematical terms:

• The derivative of a function \(f(x)\), denoted \(f'(x)\), provides a slope of the function's graph at any given point \(x\).
• Calculating a derivative involves differentiating the function concerning the variable \(x\).

In the problem exercise:- For \(f(x) = -x^3 + 1\) associated with \(x \leq 0\), the derivative is \(f'(x) = -3x^2\). This tells us how the function behaves based on changes in \(x\) value. Specifically, the function's slope decreases as it's related to a cubic graph.- For \(g(x) = -x^2 + 2x\) where \(x > 0\), the derivative \(g'(x) = -2x + 2\) is linear. This relation monitors changes giving insights on the function’s linear increase or decrease.Through derivatives, we get a clearer picture of when and how the behavior of functions transitions, essential for optimizing and understanding real-world phenomena.

Increasing and Decreasing Functions

Recognizing whether a function is increasing or decreasing depends on the sign of its derivative. These patterns can be quite illuminating for understanding the behavior of functions over certain intervals.- **Increasing Functions:** - A function is increasing if its derivative is greater than zero over an interval. This suggests a positive slope, allowing us to understand that as \(x\) grows, \(y\) values rise.- **Decreasing Functions:** - Conversely, a function is decreasing when its derivative is less than zero, revealing a negative slope. Here, as \(x\) ascends, \(y\) values fall.In the exercise:- For \(x < 0\), the derivative \(-3x^2\) yields positive values indicating an increasing nature.- For the interval \(0 < x < 1\), the derivative \(-2x + 2\) maintains positive outputs showing this part of the function is increasing.- At \(x > 1\), \(-2x + 2\) becomes negative, indicating this segment is decreasing.Understanding when a function increases and decreases helps not only in sketching precise graphs but also in identifying peak and least values—the vital aspect of calculus applications.