Problem 11 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ h(x)=-x^{3}+2 x^{2} $$

Step-by-Step Solution

Answer

The function is increasing on \((0, \frac{4}{3})\) and decreasing elsewhere. It has a local maximum (also absolute) of \( \frac{32}{27} \) at \( x = \frac{4}{3} \) and a local minimum of 0 at \( x=0 \).

Step 1: Find the derivative

To determine where the function is increasing or decreasing, we first need to find its derivative. For the function \( h(x) = -x^3 + 2x^2 \), apply the power rule:\[ h'(x) = \frac{d}{dx}(-x^3 + 2x^2) = -3x^2 + 4x. \]

Step 2: Identify critical points

Critical points occur where the derivative \( h'(x) \) is zero or undefined. Solve \( -3x^2 + 4x = 0 \):\[ x(-3x + 4) = 0 \]This gives critical points at \( x = 0 \) and \( x = \frac{4}{3} \).

Step 3: Determine intervals of increase and decrease

Create a sign chart for \( h'(x) \) around the critical points \( x = 0 \) and \( x = \frac{4}{3} \). Choose test points from intervals \((-\infty, 0)\), \((0, \frac{4}{3})\), and \((\frac{4}{3}, \infty)\):- Test \( x = -1 \): \( h'(-1) = -3(-1)^2 + 4(-1) = -3 - 4 = -7 \) (negative, hence decreasing).- Test \( x = 1 \): \( h'(1) = -3(1)^2 + 4(1) = -3 + 4 = 1 \) (positive, hence increasing).- Test \( x = 2 \): \( h'(2) = -3(2)^2 + 4(2) = -12 + 8 = -4 \) (negative, hence decreasing).Thus, the function is decreasing on \((-\infty, 0)\) and \((\frac{4}{3}, \infty)\), and increasing on \((0, \frac{4}{3})\).

Step 4: Identify local extreme values

We can find local extrema by checking the sign changes of the derivative around the critical points:- At \( x = 0 \), the derivative changes from negative to positive (decreasing to increasing), indicating a local minimum.- At \( x = \frac{4}{3} \), the derivative changes from positive to negative (increasing to decreasing), indicating a local maximum.Thus, there is a local minimum at \( x = 0 \) and a local maximum at \( x = \frac{4}{3} \).

Step 5: Evaluate local extrema

Find the function values at the critical points to determine the local extrema values:- \( h(0) = -(0)^3 + 2(0)^2 = 0 \)- \( h\left(\frac{4}{3}\right) = -\left(\frac{4}{3}\right)^3 + 2\left(\frac{4}{3}\right)^2 \)Calculate: \( h\left(\frac{4}{3}\right) = -\frac{64}{27} + \frac{32}{9} = \frac{-64}{27} + \frac{96}{27} = \frac{32}{27} \)The local minimum is 0 at \( x=0 \) and the local maximum is \( \frac{32}{27} \) at \( x = \frac{4}{3} \).

Step 6: Determine if any extremas are absolute

Since \( \lim_{x \to \pm \infty} h(x) = -\infty \), the absolute maximum cannot occur as \( x \to \infty \) or \( x \to -\infty \). The local maximum \( \frac{32}{27} \) at \( x = \frac{4}{3} \) is higher than the value at the other critical point and is therefore an absolute maximum. There is no absolute minimum because the function projectively decreases without bound as \( x \to \pm \infty \).

Step 7: Support findings with a graph

Graph the function \( h(x) = -x^3 + 2x^2 \) using a graphing utility. Verify the intervals of increase and decrease, the location of the local extremas, and confirm that the local maximum at \( x = \frac{4}{3} \) is highest. Ensure the graph supports that the function extends to \(-\infty\) on \( y \)-axis as \( x \to \pm \infty \).

Key Concepts

Derivative

The concept of a derivative is fundamental in calculus and helps to describe the behavior of functions. A derivative represents the rate at which a function is changing at any given point. Imagine you're driving a car; the speedometer shows your speed at a particular moment. Similarly, the derivative indicates how fast a function's value is changing at a specific point.

To find the derivative of a function like \( h(x) = -x^3 + 2x^2 \), we apply differentiation rules. In this case, we use the power rule, which tells us how to handle each term in the function. We multiply the coefficient by the power of \( x \), then reduce the power by one.

This results in:

• For \( -x^3 \), the derivative is \( -3x^2 \).
• For \( 2x^2 \), the derivative is \( 4x \).

Hence, the derivative of the function is \( h'(x) = -3x^2 + 4x \). This derivative will help us discover more about how the function behaves over different intervals.

Critical Points

Critical points are the places in the domain of a function where it might have a high point (a peak) or a low point (a valley). These points are crucial because they're where the function's rate of change switches from increasing to decreasing or vice versa.

To find critical points, we solve the equation \( h'(x) = 0 \) since critical points occur where the derivative is zero or undefined. Remember, plugging in the value of \( x \) should give zero for the derivative:

• So, for the function \( h(x) = -x^3 + 2x^2 \), we solve \( -3x^2 + 4x = 0 \).
• Factoring gives us \( x(-3x + 4) = 0 \). Thus, the solutions are \( x = 0 \) and \( x = \frac{4}{3} \).

These points, \( x = 0 \) and \( x = \frac{4}{3} \), are the critical points that we will analyze further to understand the function's behavior.

Local Extrema

Local extrema are the peaks and valleys within a particular segment of the function's domain. They can be either a local maximum, where the function reaches the highest point in that neighborhood, or a local minimum, where it plunges to the lowest point.

We find these by observing how the derivative changes sign at a function's critical points:

• If the derivative changes from negative to positive at a critical point, there's a local minimum (a dip).
• If it changes from positive to negative, there's a local maximum (a peak).

In our function, at \( x = 0 \), the derivative changes from negative to positive, indicating a local minimum.

At \( x = \frac{4}{3} \), it changes from positive to negative, signaling a local maximum. Hence, we have:

• Local minimum at \( x = 0 \), value \( h(0) = 0 \).
• Local maximum at \( x = \frac{4}{3} \), value \( h\left(\frac{4}{3}\right) = \frac{32}{27} \).

These points give us a clearer picture of where the function "turns" its behavior in a specific range.

Absolute Extrema

Absolute extrema represent the highest and lowest points of the function over the entire domain. Unlike local extrema, they aren't restricted to the neighborhoods of the critical points. They consider the function's behavior across all possible input values.

To find absolute extrema, we consider both the critical points and the behavior of the function as \( x \to \pm\infty \). For our function:

• As \( x \to \pm\infty \), \( h(x) \to -\infty \) because of the \( -x^3 \) term dominating negative growth.
• The highest value within our critical points is therefore an absolute maximum.

The function has:

• No absolute minimum, as it decreases without bound on both ends.
• An absolute maximum at \( x = \frac{4}{3} \) with the value \( \frac{32}{27} \).

This analysis confirms that the function achieves its peak value at the critical point \( x = \frac{4}{3} \), and no lower bound exists due to the function's unbounded decline.