Problem 227 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

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

Step 1: Find the derivative of f(x)

To determine the intervals where the function is increasing or decreasing, we first find the first derivative of the function, \( f(x) = x^{11} - 6x^{10} \). The derivative is found using the power rule: \( f'(x) = 11x^{10} - 60x^9 \).

Step 2: Solve f'(x) = 0 for critical points

To find the critical points, set \( f'(x) = 0 \). \( 11x^{10} - 60x^9 = 0 \) simplifies to \( x^9(11x - 60) = 0 \). This gives critical points at \( x = 0 \) and \( x = \frac{60}{11} \).

Step 3: Test intervals for increasing/decreasing

Use a sign chart or test points around the critical points to determine the intervals of increase or decrease. For \( x < 0 \), \( f'(x) > 0 \), thus \( f \) is increasing. In \( 0 < x < \frac{60}{11} \), \( f'(x) < 0 \), so \( f \) is decreasing. For \( x > \frac{60}{11} \), \( f'(x) > 0 \), leading to \( f \) increasing.

Step 4: Determine local minima and maxima

Since \( f \) changes from increasing to decreasing at \( x = 0 \), it has a local maximum at this point. At \( x = \frac{60}{11} \), \( f \) changes from decreasing to increasing, indicating a local minimum.

Step 5: Find the second derivative f''(x)

To analyze concavity, find the second derivative: \( f''(x) = 110x^9 - 540x^8 \). Factor to get \( f''(x) = 10x^8(11x - 54) \).

Step 6: Solve f''(x) = 0 for possible inflection points

Set \( 10x^8(11x - 54) = 0 \). Solve to find possible inflection points at \( x = 0 \) or \( x = \frac{54}{11} \).

Step 7: Test intervals for concavity changes

Determine concavity by testing the sign of \( f''(x) \) in the intervals. For \( x < 0 \), \( f''(x) < 0 \), indicating concave down. In \( 0 < x < \frac{54}{11} \), \( f''(x) > 0 \), concave up. For \( x > \frac{54}{11} \), \( f''(x) < 0 \), again concave down.

Step 8: Confirm inflection points

Inflection points occur where concavity changes. Thus, \( x = 0 \) and \( x = \frac{54}{11} \) are confirmed as points of inflection.

Key Concepts

Derivative

A derivative in calculus represents how a function changes as its input changes. It tells us the rate of change or the slope of the function at any given point. Calculating derivatives is a fundamental skill in calculus. To find a function's derivative, we use differentiation rules. In this exercise, we used the power rule to differentiate the function.The power rule states: for any function of the form \( ax^n \), the derivative is given by \( nax^{n-1} \). Applying this rule to the function \( f(x) = x^{11} - 6x^{10} \), we find the derivative:

• \( \frac{d}{dx}[x^{11}] = 11x^{10} \)
• \( \frac{d}{dx}[-6x^{10}] = -60x^9 \)

Combining these results gives us the first derivative: \( f'(x) = 11x^{10} - 60x^9 \). This derivative helps us determine where the function is increasing or decreasing by finding its critical points.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are crucial in identifying potential maxima or minima of the function. To find the critical points for the function \( f(x) = x^{11} - 6x^{10} \), we set the derivative equal to zero:\[ 11x^{10} - 60x^9 = 0 \]Factoring out \( x^9 \) gives us:\[ x^9(11x - 60) = 0 \]Solving for \( x \) results in two critical points:

• \( x = 0 \)
• \( x = \frac{60}{11} \)

These critical points indicate where we need to further examine the function to determine if they are local maxima, minima, or neither. By testing intervals around these points, we can understand the behavior of the function.

Concavity

Concavity describes how a function curves. A function is concave up where it opens upwards like a cup, and concave down where it opens downwards like a cap. To determine the concavity of a function, we use the second derivative. For our function \( f(x) \), the second derivative is calculated as:\[ f''(x) = 110x^9 - 540x^8 \]Factoring gives us:\[ f''(x) = 10x^8(11x - 54) \]To find where the function changes concavity, we assess the sign of \( f''(x) \) over different intervals of \( x \):

• For \( x < 0 \), \( f''(x) < 0 \) and the function is concave down.
• For \( 0 < x < \frac{54}{11} \), \( f''(x) > 0 \) and the function is concave up.
• For \( x > \frac{54}{11} \), \( f''(x) < 0 \) and the function is concave down again.

This analysis helps us find where the function curves and assists in locating inflection points.

Inflection Points

Inflection points are where a function changes its concavity from up to down or vice versa. These points are pivotal in understanding the shape of the graph. Inflection points occur where the second derivative is zero or changes sign.For the function \( f(x) = x^{11} - 6x^{10} \), the second derivative \( f''(x) = 10x^8(11x - 54) \) was set to zero to find possible inflection points:\[ 10x^8(11x - 54) = 0 \]Solving this yields two points:

• \( x = 0 \)
• \( x = \frac{54}{11} \)

These points indicate where the function's concavity changes, marking them as inflection points. Such points are useful for sketching the curve of the function accurately, adding depth to our understanding of its behavior.