Problem 227 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f \text { , }} \\\ {\text { c. intervals where } f \text { is concave up and concave }} \\\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . }}\end{array} $$ $$ f(x)=x^{11}-6 x^{10} $$

Step-by-Step Solution

Answer

a. Increasing on \((-\infty, 0) \cup (\frac{60}{11}, \infty)\); decreasing on \((0, \frac{60}{11})\). b. Local max at \( x = 0 \); local min at \( x = \frac{60}{11} \). c. Concave up on \((-\infty, 0) \cup (\frac{54}{11}, \infty)\); concave down on \((0, \frac{54}{11})\). d. Inflection points at \( x = 0 \) and \( x = \frac{54}{11} \).

Step 1: Find the first derivative

To determine the intervals where the function is increasing or decreasing, we first find the first derivative of the function. Given that \( f(x)=x^{11}-6x^{10} \), the derivative is calculated as follows: \( f'(x) = \frac{d}{dx}(x^{11} - 6x^{10}) = 11x^{10} - 60x^9 \).

Step 2: Find critical points

Critical points occur where the first derivative \( f'(x) \) is either zero or undefined. Set \( 11x^{10} - 60x^9 = 0 \) and solve for \( x \): \( x^9(11x - 60) = 0 \). The solutions are \( x = 0 \) and \( x = \frac{60}{11} \).

Step 3: Test intervals around critical points for increasing or decreasing behavior

Select test points from the intervals \((-\infty, 0)\), \((0, \frac{60}{11})\), and \((\frac{60}{11}, \infty)\). Calculate \( f'(x) \) for each interval:- For \( x < 0 \), use \( x = -1 \): \( f'(-1) = 11(-1)^{10} - 60(-1)^9 > 0 \). - For \( x \in (0, \frac{60}{11}) \), use \( x = 1 \): \( f'(1) = 11(1)^{10} - 60(1)^9 = -49 \) (negative).- For \( x > \frac{60}{11} \), use \( x = 10 \): \( f'(10) = 11(10)^{10} - 60(10)^9 > 0 \).Thus, \( f(x) \) is increasing on \((-\infty, 0)\) and \((\frac{60}{11}, \infty)\); decreasing on \((0, \frac{60}{11})\).

Step 4: Determine local minima and maxima

Local minima and maxima occur at critical points where \( f'(x) \) changes sign. At \( x = 0 \), \( f'(x) \) changes from positive to negative indicating a local maximum. At \( x = \frac{60}{11} \), \( f'(x) \) changes from negative to positive indicating a local minimum.

Step 5: Find the second derivative

To find intervals of concavity and inflection points, compute the second derivative. \( f''(x) = \frac{d}{dx}(11x^{10} - 60x^9) = 110x^9 - 540x^8 \).

Step 6: Find possible inflection points

Inflection points occur where the second derivative changes sign. Set \( f''(x) = 0 \): \( 110x^9 - 540x^8 = 0 \). Factor to find \( x \): \( x^8(110x - 540) = 0 \). Solutions are \( x = 0 \) and \( x = \frac{54}{11} \).

Step 7: Test intervals around possible inflection points for concavity

Select test points from intervals \((-\infty, 0)\), \((0, \frac{54}{11})\), and \((\frac{54}{11}, \infty)\). Calculate \( f''(x) \):- For \( x < 0 \), use \( x = -1 \): \( f''(-1) = 650\) (positive, concave up).- For \( x \in (0, \frac{54}{11}) \), use \( x = 1 \): \( f''(1) = -430 \) (negative, concave down).- For \( x > \frac{54}{11} \), use \( x = 10 \): \( f''(10) > 0 \) (positive, concave up).Thus, \( f(x) \) is concave up on \((-\infty, 0)\) and \((\frac{54}{11}, \infty)\); concave down on \((0, \frac{54}{11})\). Inflection points are at \( x = 0 \) and \( x = \frac{54}{11} \).

Key Concepts

Increasing and Decreasing Intervals

When determining where a function is increasing or decreasing, we rely on its first derivative. The function is increasing in intervals where this derivative is positive and decreasing where it is negative.
Let's take a deeper dive into why this is the case:

• An increasing function means that as you move from left to right on the graph, the value of the function rises. This corresponds to a positive slope on the graph, which is evidenced by a positive first derivative.
• Conversely, a decreasing function has a downward slope, corresponding to a negative derivative.

To find these intervals for a function like \(f(x) = x^{11} - 6x^{10}\):

• First, compute the derivative \(f'(x) = 11x^{10} - 60x^9\).
• Next, identify the critical points, where \(f'(x) = 0\), to find changes in the function's direction.
• Lastly, test intervals between these critical points to determine whether \(f'(x)\) is positive or negative in these intervals.

For our function, \(f(x)\) is increasing on \((-\infty, 0)\) and \((60/11, \infty)\); it's decreasing on \((0, 60/11)\). In these intervals, the behavior of \(f'(x)\) tells us whether \(f(x)\) is climbing or falling.

Local Minima and Maxima

Local minima and maxima are critical points where the function reaches a minimum or maximum value within a surrounding interval. This happens when the first derivative changes its sign.

• A local maximum occurs at a point where \(f'(x)\) changes from positive to negative. It is akin to a peak in the function.
• A local minimum occurs at a point where \(f'(x)\) changes from negative to positive, resembling a valley in the graph.

Using these principles with the function \(f(x) = x^{11} - 6x^{10}\), we can find:

• At \(x = 0\), \(f'(x)\) switches from positive to negative, marking a local maximum.
• At \(x = \frac{60}{11}\), \(f'(x)\) switches from negative to positive, denoting a local minimum.

These points provide the highest or lowest surrounding values, but not necessarily the absolute highest or lowest on the entire graph.

Concavity and Inflection Points

Concavity relates to how the curve of the graph bends, and it is determined by the second derivative. A function's graph can be either 'concave up,' like a U, or 'concave down,' like an upside-down U.

• When the second derivative \(f''(x) > 0\), the function is concave up, and the graph appears like a smile.
• When \(f''(x) < 0\), the function is concave down, and the graph looks like a frown.

To find the inflection points, where the concavity changes:

• Calculate the second derivative \(f''(x)\) of the function.
• Set \(f''(x) = 0\) to find potential points of inflection.
• Test intervals around these points to determine where the sign changes, indicating a change in concavity.

For \(f(x) = x^{11} - 6x^{10}\):

• \(f(x)\) is concave up on \((-\infty, 0)\) and \((54/11, \infty)\).
• It is concave down on \((0, 54/11)\).

The points \(x = 0\) and \(x = 54/11\) are inflection points, where the concavity changes.

Derivatives and Critical Points

Derivatives give us powerful tools to analyze functions. The first derivative \(f'(x)\) tells us about the slope or rate of change of the function. It helps in identifying critical points, where the function's behavior might change.

• Critical points occur where \(f'(x)\) is zero or undefined. At these points, the function might have a local minimum, maximum, or neither.
• The role of the critical points is crucial in understanding the function's overall graph and behavior.

In our analysis of \(f(x) = x^{11} - 6x^{10}\):

• The first derivative is \(f'(x) = 11x^{10} - 60x^9\). By setting the derivative to zero, we find critical points at \(x = 0\) and \(x = \frac{60}{11}\).
• These points are where \(f(x)\) could switch from increasing to decreasing or vice versa.

Understanding derivatives and critical points helps us predict and explain significant "landmarks" within a function's graph.