Problem 21 - University Calculus: Early Transcendentals

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$h(x)=-x^{3}+2 x^{2}$$

Step-by-Step Solution

Answer

The function is increasing on \((0, \frac{4}{3})\) and decreasing on \((-\infty, 0)\) and \((\frac{4}{3}, \infty)\). Local minimum at \(x = 0\), and local maximum \(\frac{32}{27}\) at \(x = \frac{4}{3}\).

Step 1: Find the first derivative

The first step to determine where the function is increasing or decreasing is to find its first derivative. For the function \( h(x) = -x^3 + 2x^2 \), we use the power rule for derivatives. Thus, the first derivative is:\[h'(x) = (-3x^2) + (4x) = -3x^2 + 4x\]

Step 2: Set the derivative equal to zero

Next, we find the critical points by setting \( h'(x) = 0 \) and solving for \( x \).\[-3x^2 + 4x = 0\]Factor out the common term \( x \):\[x(-3x + 4) = 0\]This gives us the critical points at \( x = 0 \) and \( x = \frac{4}{3} \).

Step 3: Determine intervals for increasing/decreasing behavior

Use the critical points to test intervals in the first derivative to find where the function is increasing or decreasing. We will pick test points in the intervals \((-\infty, 0)\), \((0, \frac{4}{3})\), and \((\frac{4}{3}, \infty)\).- For \( x < 0 \), let \( x = -1 \): \[h'(-1) = -3(-1)^2 + 4(-1) = -3 - 4 = -7 \] (negative, so decreasing)- For \( 0 < x < \frac{4}{3} \), let \( x = 1 \): \[h'(1) = -3(1)^2 + 4(1) = -3 + 4 = 1 \] (positive, so increasing)- For \( x > \frac{4}{3} \), let \( x = 2 \): \[h'(2) = -3(2)^2 + 4(2) = -12 + 8 = -4 \] (negative, so 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 extrema by testing critical points

To find local maximums or minimums, test the critical points using the first derivative test.- At \( x = 0 \), the derivative changes from negative to positive, indicating a local minimum.- At \( x = \frac{4}{3} \), the derivative changes from positive to negative, indicating a local maximum.The local minimum value of the function is at \( x = 0 \) and the local maximum value of the function is at \( x = \frac{4}{3} \).

Step 5: Calculate local extrema values

Substitute the critical points back into the original function to find the values of the local extrema.- For \( x = 0 \): \[h(0) = -(0)^3 + 2(0)^2 = 0\]- For \( x = \frac{4}{3} \): \[h\left(\frac{4}{3}\right) = -\left(\frac{4}{3}\right)^3 + 2\left(\frac{4}{3}\right)^2 = -\frac{64}{27} + \frac{32}{9}\]Convert \( \frac{32}{9} \) to \( \frac{96}{27} \):\[-\frac{64}{27} + \frac{96}{27} = \frac{32}{27}\]Thus, the local maximum value is \( \frac{32}{27} \) at \( x = \frac{4}{3} \), and the local minimum value is 0 at \( x = 0 \).

Key Concepts

Derivative

In calculus, the concept of a derivative is fundamental to understanding how functions change. The derivative essentially provides us a tool to measure how a function's output value changes with respect to changes in its input value. For a given function, the derivative at a particular point indicates the rate of change or the slope of the tangent line to the function at that point.
For the function \( h(x) = -x^3 + 2x^2 \), we begin by deriving its first derivative, which requires applying the power rule. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Hence, for our function, the first derivative is:

• \( h'(x) = -3x^2 + 4x \)

This derivative provides a new function that helps us determine the behavior of the original function by indicating the slope at any given point.

Critical Points

Critical points are specific input values of a function where the derivative is zero or undefined. These points are essential because they often indicate where crucial changes occur in the function's behavior, such as changes from increasing to decreasing or vice-versa.
To find the critical points of our function \( h(x) \), we set its first derivative to zero: \(-3x^2 + 4x = 0\). Solving this equation requires factoring:

• \( x(-3x + 4) = 0 \)

From this, we find two critical points: \( x = 0 \) and \( x = \frac{4}{3} \).
These points divide the number line into intervals where the function's increasing or decreasing trend can be tested with test values.

Increasing/Decreasing Intervals

Identifying where a function is increasing or decreasing is essential to understanding its overall behavior. A function is increasing on an interval if its derivative is positive for every point in that interval. Similarly, it is decreasing if the derivative is negative.
For \( h'(x) = -3x^2 + 4x \), we test the sign of the derivative in different intervals that are determined by the critical points, \( x = 0 \) and \( x = \frac{4}{3} \).
By choosing test values in each interval, we find:

• Decreasing on \((-fty, 0)\) since \( h'(-1) = -7 \) (a negative number).
• Increasing on \((0, \frac{4}{3})\) because \( h'(1) = 1 \) (a positive number).
• Decreasing on \((\frac{4}{3}, fty)\) with \( h'(2) = -4 \) (a negative number).

This analysis shows us where the function rises and falls.

Local Extrema

Local extrema refer to the highest or lowest points on a specific interval of a function. These are points where a function changes direction from increasing to decreasing or vice-versa. Local extrema are found at critical points where the derivative changes sign.
For \( h(x) \), use the first derivative test as follows:

• At \( x = 0 \), the derivative changes from negative to positive, indicating a local minimum.
• At \( x = \frac{4}{3} \), it switches from positive to negative, indicating a local maximum.

To find the actual values of these extrema, substitute back into the original function.

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

Knowing these extrema points helps us understand potential highest or lowest outputs within particular sections of the function.