Problem 21 - Thomas Calculus

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 \((-\infty, \frac{4}{3})\) and decreasing on \((\frac{4}{3}, \infty)\). It has a local and absolute maximum of \(\frac{32}{27}\) at \(x = \frac{4}{3}\).

Step 1: Find the derivative of the function

We start by finding the derivative of the function \( h(x) = -x^3 + 2x^2 \). The derivative \( h'(x) \) gives us the rate of change of the function, which will help determine where the function is increasing or decreasing.The derivative is calculated as follows:\[h'(x) = \frac{d}{dx}(-x^{3} + 2x^2) = -3x^2 + 4x\]

Step 2: Find the critical points

To find the critical points, we set the derivative \( h'(x)=0 \) and solve for \( x \):\[-3x^2 + 4x = 0\]Factor out \( x \):\[x(-3x + 4) = 0\]This gives the critical points: \( x = 0 \) and \( x = \frac{4}{3} \).

Step 3: Determine intervals of increase and decrease

To determine the intervals, examine the sign of \( h'(x)\) on intervals between and beyond the critical points \( x = 0 \) and \( x = \frac{4}{3} \):* For \( x < 0 \), choose \( x = -1 \), \( h'(-1) = 3 + 4 = 7 \) (positive, increasing).* For \( 0 < x < \frac{4}{3} \), choose \( x = 1 \), \( h'(1) = -3 + 4 = 1 \) (positive, increasing).* For \( x > \frac{4}{3} \), choose \( x = 2 \), \( h'(2) = -12 + 8 = -4 \) (negative, decreasing).Thus, the function is increasing on \((-\infty, \frac{4}{3})\) and decreasing on \((\frac{4}{3}, \infty)\).

Step 4: Identify local extrema

Examine the critical points for local extrema:- At \( x = 0 \), the function moves from increasing to increasing; thus, there is no local extremum.- At \( x = \frac{4}{3} \), the function moves from increasing to decreasing, indicating a local maximum.Calculate \( h \left( \frac{4}{3} \right) \):\[h \left( \frac{4}{3} \right) = -\left( \frac{4}{3} \right)^3 + 2 \left( \frac{4}{3} \right)^2 = -\frac{64}{27} + \frac{32}{9} = \frac{32}{27}\]Thus, a local maximum of \( \frac{32}{27} \) occurs at \( x = \frac{4}{3} \).

Step 5: Check endpoints and asymptotic behavior for absolute extrema

Since the function and derivative are defined for all \( x \) and go to \(-\infty \) as \( x \to \infty \) and \( x \to -\infty \), the local maximum at \( x = \frac{4}{3} \) is the highest point the function reaches, making it also the absolute maximum.There is no absolute minimum, as \( h(x) \to -\infty \) as \( x \to \pm \infty \).

Key Concepts

Derivative

The derivative of a function is essentially the rate at which the function changes with respect to one of its variables. For our function, \[ h(x) = -x^3 + 2x^2, \] the derivative is \[ h'(x) = -3x^2 + 4x. \] The process of finding a derivative involves applying rules such as the power rule, which states that the derivative of \( ax^n \) is \( n \cdot ax^{n-1} \). Here, the derivative helps us determine how the function behaves—specifically, identifying where it increases or decreases. By setting the derivative to zero \( h'(x) = 0 \), we can find the critical points where potential maxima, minima, or points of inflection might occur. This is because these are the points at which the change in the function disappears briefly.

Intervals of Increase and Decrease

To determine where a function is increasing or decreasing, we look at the sign of its derivative. For our function, we calculated its derivative as \( h'(x) = -3x^2 + 4x \). By testing regions around critical points, we establish where the function is increasing or decreasing.
If \( h'(x) > 0 \) for some interval, the function is increasing in that interval. Conversely, if \( h'(x) < 0 \) for some interval, the function is decreasing there.

• For \( x < 0 \), \( h'(x) \) is positive, meaning the function is increasing.
• Between \( x = 0 \) and \( x = \frac{4}{3} \), the derivative is still positive, so the function continues to increase.
• For \( x > \frac{4}{3} \), \( h'(x) \) is negative, indicating the function is decreasing.

Thus, the function increases on \((-\infty, \frac{4}{3})\) and decreases on \((\frac{4}{3}, \infty)\). Checking these signs gives us a clear map of where the function is actively rising or falling.

Local Extrema

Local extrema are the peaks and valleys in the function's graph. These can be either a local maximum or a local minimum. At these points, the function will change direction from increasing to decreasing or vice versa.
We can identify local extrema by looking at critical points—points where the derivative is zero or undefined. In our case, critical points appear at \( x = 0 \) and \( x = \frac{4}{3} \).
Only the point \( x = \frac{4}{3} \) presents a meaningful change from increasing to decreasing, exhibiting a local maximum. At this point, \[ h \left( \frac{4}{3} \right) = \frac{32}{27}. \] At \( x = 0 \), the function continues increasing, meaning there's no local minimum or maximum.

Absolute Extrema

Absolute extrema are the highest and lowest points a function reaches over its entire domain. Given our function's behavior, we seek to determine not only local behavior but its overall performance.
For a function like \[ h(x) = -x^3 + 2x^2, \] which extends indefinitely, absolute extrema are particularly significant. As \( x \to \infty \) or \( x \to -\infty \), the function heads towards \(-\infty\), indicating there is no absolute minimum.
However, because we identified a local maximum at \( x = \frac{4}{3} \) and have no higher points across the entire function, this point becomes the absolute maximum with a value of \[ \frac{32}{27}. \] This thorough examination shows how the local peak is indeed the overall highest point the function achieves.