Problem 57 - Calculus and its applications

Question

Graph each function. Then determine any critical values, inflection points, intervals over which the function is increasing or decreasing, and the concavity. $$ f(x)=e^{(1 / 3) x} $$

Step-by-Step Solution

Answer

The function \( f(x) = e^{(1/3)x} \) is increasing and concave up on \( (-\infty, \infty) \) with no critical or inflection points.

Step 1: Graph the Function

First, recognize that the function given is an exponential function. The graph of \( f(x) = e^{(1/3)x} \) will have a y-intercept at \( y = 1 \) when \( x = 0 \), consistent with the point \( (0,1) \). The graph will increase as \( x \) becomes larger because the base, \( e \), raised to any positive power becomes larger.

Step 2: Find the Critical Values

Critical values occur where the first derivative equals zero or is undefined. The first derivative of \( f(x) = e^{(1/3)x} \) is \( f'(x) = \frac{1}{3} e^{(1/3)x} \). \( f'(x) \) never equals zero and is always defined since \( e^{x} \) and its derivatives are always positive. Therefore, there are no critical values.

Step 3: Determine Increasing and Decreasing Intervals

Since \( f'(x) = \frac{1}{3} e^{(1/3)x} \) is positive for all \( x \), the function is increasing on its entire domain, \( (-\infty, \infty) \).

Step 4: Calculate the Second Derivative

Find the second derivative to determine concavity and inflection points: \( f''(x) = \frac{1}{9} e^{(1/3)x} \).

Step 5: Determine Concavity and Inflection Points

Since \( f''(x) = \frac{1}{9} e^{(1/3)x} \) is positive for all \( x \), the function is concave up on its entire domain, \( (-\infty, \infty) \). There are no inflection points because there is no change in concavity.

Key Concepts

Exponential Function

Exponential functions are quite fascinating and form an essential part of calculus graph analysis. In our example, the function given is \( f(x) = e^{(1/3)x} \). Here, \( e \) stands for Euler's number, approximately equal to 2.71828. An exponential function is characterized by a constant base, raised to a variable power. This results in a rapid increase or decrease, depending on the exponent's sign.
A couple of key features of exponential functions include:

• The base \( e \) is always positive, which means the function will never intercept the x-axis.
• Exponential functions with a positive exponent value grow as \( x \) increases.

In the function \( f(x) = e^{(1/3)x} \), as \( x \) increases from negative to positive infinity, \( f(x) \) enhances from 0 upwards, displaying its increasing nature.

Critical Values

Critical values are crucial in identifying where a function changes its direction from increasing to decreasing or vice versa. This typically occurs where the first derivative of the function equals zero or is undefined.
For \( f(x) = e^{(1/3)x} \), the first derivative \( f'(x) = \frac{1}{3} e^{(1/3)x} \) tells us about the rate of change. Since \( e^{(1/3)x} \) is always positive, the derivative is also always positive and never zero.
This implies that \( f(x) \) has no critical values, meaning there are no turning points where the function stops increasing or starts decreasing.

Increasing and Decreasing Intervals

Understanding the intervals over which the function increases or decreases relies on the sign of its first derivative. With \( f(x) = e^{(1/3)x} \), the first derivative \( f'(x) = \frac{1}{3} e^{(1/3)x} \) is positive for all real numbers \( x \).
This positivity of the derivative indicates that the function is always increasing, with no intervals where it decreases.
Thus, \( f(x) \) is increasing on its entire domain, \((-\infty, \infty)\), affirming it accelerates upwards without any descending behavior.

Concavity and Inflection Points

The concept of concavity describes whether a function curves upwards or downwards. Inflection points are the places where the curve changes its concavity direction.
To find concavity, we examine the second derivative. For \( f(x) = e^{(1/3)x} \), the second derivative is \( f''(x) = \frac{1}{9} e^{(1/3)x} \).
Since \( f''(x) \) is positive for every \( x \), the function is concave up across its whole domain, \((-\infty, \infty)\). A concave up function resembles a "U"-shape, continuously bending upwards.
Having no sign change in the second derivative implies that there are no inflection points in this function. The lack of inflection points tells us that \( f(x) \) maintains its upward curve without switching to a downwards direction.