Problem 35 Sketch the graph of each functio... [FREE SOLUTION]

91影视

Calculus and Its Applications

Marvin L. Bittinger, David J. Ellenbogen, Scott A. Surgent

$Math Studyset 91影视 Explanations$ Math

10 Edition

Chapter 2: Problem 35

Sketch the graph of each function. List the coordinates of where extrema or points of inflection occurs State where the function is increasing or decreasing, as well as where it is concave up or concave down. $$f(x)=(x+1)^{2 / 3}$$

Short Answer

Expert verified

The point of inflection is at $(-1, 0)$. The function decreases for $x < -1$ and increases for $x > -1$. It is concave down everywhere.

Step by step solution

- Determine the Domain and Range

The function is defined as $f(x) = (x+1)^{2/3}$.The domain of this function is all real numbers, since any real number can be plugged into the function without causing any issues. The range will also include all non-negative numbers.

- Identify Critical Points

Differentiate the function to find points where the function's derivative is zero or undefined. First, note that:\[ f(x) = (x+1)^{2/3} \]The derivative is:\[ f'(x) = \frac{2}{3}(x+1)^{-1/3} \]The derivative is undefined at $ x = -1 $. This point is a candidate for a critical point.

- Determine Extrema and Points of Inflection

Evaluate the second derivative for points of inflection:\[ f''(x) = \frac{d}{dx}( \frac{2}{3}(x+1)^{-1/3} ) = \frac{-2}{9}(x+1)^{-4/3} \]The second derivative is undefined at $ x = -1 $. This indicates a point of inflection.

- Analyze Increasing/Decreasing Intervals

Use the first derivative analysis:- For $x > -1$, $ f'(x) > 0 $, so the function is increasing.- For $x < -1$, $ f'(x) < 0 $, so the function is decreasing.

- Analyze Concavity

Use the second derivative to determine concavity:- For $ x > -1 $, $ f''(x) < 0 $, so the function is concave down.- For $ x < -1 $, $ f''(x) < 0 $, so the function is also concave down in this interval.

- Sketch the Graph

Based on the information collected:1. The function has a critical point and point of inflection at $ x = -1 $.2. The function is decreasing for $ x < -1 $ and increasing for $ x > -1 $.3. The function is concave down across its entire domain.Sketch the graph integrating the critical point, increasing/decreasing intervals, and concavity characteristics.

- List Coordinates of Extrema and Points of Inflection

The point of inflection is at $ (-1, 0) $. There are no local extrema because the function continuously decreases before $ x = -1 $ and continuously increases after $ x = -1 $.

- State Intervals of Increase/Decrease and Concavity

The function is increasing for $ x > -1 $ and decreasing for $ x < -1 $. The function is concave down over its entire domain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points

Critical points are places on the graph where the derivative is either zero or undefined. These points often signify where a function changes from increasing to decreasing or vice versa. For the function $ f(x) = (x+1)^{2/3} $, we found the first derivative as:
\[ f'(x) = \frac{2}{3}(x+1)^{-1/3} \]
Since $ f'(x) $ is undefined at $ x = -1 $, this is a critical point. This critical point tells us important information about the behavior of the function.

Increasing and Decreasing Intervals

To find where a function is increasing or decreasing, we look at its first derivative. A function is increasing where its first derivative is positive and decreasing where its first derivative is negative. For $ f(x) = (x+1)^{2/3} $, the first derivative $ f'(x) = \frac{2}{3}(x+1)^{-1/3} $ reveals:

For $ x > -1 $, $ f'(x) > 0 $, meaning the function is increasing.
For $ x < -1 $, $ f'(x) < 0 $, meaning the function is decreasing.

Understanding these intervals helps us sketch the function more accurately and understand how it behaves across different values of $ x $.

Concavity

Concavity describes the direction the graph bends. A function is concave up when its second derivative is positive and concave down when its second derivative is negative. For $ f(x) = (x+1)^{2/3} $, the second derivative is:
\[ f''(x) = \frac{-2}{9}(x+1)^{-4/3} \]

For $ x > -1 $, $ f''(x) < 0 $, so the function is concave down.
For $ x < -1 $, $ f''(x) < 0 $, so it also remains concave down.

This consistent concave down behavior across all $ x $ shows that the function never bends upwards.

Point of Inflection

A point of inflection is where the concavity of a function changes. For $ f(x) = (x+1)^{2/3} $, we determined the second derivative:
\[ f''(x) = \frac{-2}{9}(x+1)^{-4/3} \]
The second derivative is undefined at $ x = -1 $, indicating a potential point of inflection. However, since the function is concave down before and after this point, $ x = -1 $ is the point where the concavity transitions without changing signs. Therefore, the point $ (-1, 0) $ is crucial in understanding the function's curvature.

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