Problem 62 - Calculus Early Transcendentals

Question

More graphing Sketch a complete graph of the following functions. Use analytical methods and a graphing utility together in a complementary way. $$f(x)=3 \sqrt[4]{x}-\sqrt{x}-2$$

Step-by-Step Solution

Answer

Question: Sketch a graph of the function $$f(x)=3\sqrt[4]{x}-\sqrt{x}-2$$. Answer: To sketch the graph of the function, determine that the domain is $$x \geq 0$$. Note that the function is increasing and concave up on its entire domain, with no critical points or points of inflection. There are also no asymptotes. To confirm these findings and obtain the exact shape of the graph, use a graphing calculator or online graphing utility.

Step 1: Determine the domain of the function

First, we need to find the domain of the function, which is the set of all possible x-values for which the function is defined. Since the two square root expressions are defined only when the input x is non-negative and has real values, the domain of this function is: $$x \geq 0$$

Step 2: Find the first and second derivatives

We will find the first and second derivatives of the function to analyze its critical points and points of inflection. First derivative: $$f'(x) = \frac{3}{4\sqrt[4]{x^3}} - \frac{1}{2\sqrt{x}}$$ Second derivative: $$f''(x) = -\frac{9}{16 \sqrt{x^7}} + \frac{1}{4 \sqrt{x^3}}$$

Step 3: Critical points, and increasing and decreasing intervals

To find the critical points of the function, we set the first derivative equal to zero and solve for x: $$f'(x) = 0$$ We find that there are no critical points for x ≥ 0. To identify the increasing and decreasing intervals of the function, we examine the sign of the first derivative for different values of x in the domain (x ≥ 0). Since $$f'(x) > 0$$ for all x in the domain, the function is increasing on its entire domain.

Step 4: Points of inflection, and concavity

To find the points of inflection, we set the second derivative equal to zero and solve for x: $$f''(x) = 0$$ There are no points of inflection for x ≥ 0. To identify the concavity of the function, we examine the sign of the second derivative for different values of x in the domain (x ≥ 0). Since $$f''(x) > 0$$ for all x in the domain, the function is concave up on its entire domain.

Step 5: Asymptotes and Limits

As there is no division by zero and the function is defined for all x in the domain, there are no vertical asymptotes. The function does not show any particular behavior as it approaches infinity, so there are no horizontal asymptotes.

Step 6: Use a graphing utility

To confirm the results and obtain the exact shape of the graph of the function, use a graphing calculator or an online graphing utility like Desmos or GeoGebra. After plotting the function on a graphing utility, you should see that the function is increasing and concave up on its entire domain, as determined analytically. There are no critical points or points of inflection, and there are no asymptotes.

Key Concepts

Domain of Functions

The domain of a function refers to the complete set of possible values of the independent variable, usually represented by "x," for which the function is defined and produces real numbers. In the function given, \( f(x)=3 \sqrt[4]{x}-\sqrt{x}-2 \), we observe two terms involving roots: \( \sqrt[4]{x} \) and \( \sqrt{x} \). These expressions require the input \( x \) to be non-negative in order to result in real values. Consequently, the domain of this function is limited to non-negative numbers. Thus, the domain is defined as:

• \( x \geq 0 \)

Understanding the domain is crucial before diving into further analysis of a function, as it sets the boundaries for all subsequent calculations.

First and Second Derivatives

Derivatives are essential tools in calculus that help determine the rate at which a function changes. The first derivative of a function, denoted as \( f'(x) \), provides information on the function's slope or rate of change. For functions involving roots, derivatives can highlight interesting behavior patterns.For \( f(x) = 3 \sqrt[4]{x} - \sqrt{x} - 2 \), the first derivative is:

• \( f'(x) = \frac{3}{4\sqrt[4]{x^3}} - \frac{1}{2\sqrt{x}} \)

This equation helps examine how fast the function is increasing or decreasing. The second derivative \( f''(x) \) focuses on how the rate of change itself changes, revealing concavity:

• \( f''(x) = -\frac{9}{16 \sqrt{x^7}} + \frac{1}{4 \sqrt{x^3}} \)

Analyzing these derivatives provides valuable insights into the function's geometry or shape.

Critical Points and Intervals

The first derivative is also used to find critical points and analyze the increasing or decreasing nature of the function. Critical points occur where the derivative equals zero or is undefined, potentially marking locations of a peak, valley, or plateau in the function.For the function \( f(x)=3 \sqrt[4]{x}-\sqrt{x}-2 \), finding \( f'(x)=0 \) reveals no critical points for \( x \geq 0 \). Calculating and observing the sign of the first derivative throughout the domain, we find:

• Since \( f'(x) > 0 \), the function is increasing for all \( x \geq 0 \)

No critical points suggest the function doesn't change direction within the domain, maintaining a consistent behavior pattern across all defined x-values.

Concavity and Inflection Points

Concavity describes whether a function's curve is opening upwards or downwards. Analyzing the second derivative, we determine concavity and locate possible inflection points where the concavity changes.For \( f(x)=3 \sqrt[4]{x}-\sqrt{x}-2 \), the second derivative is examined for its sign:

• \( f''(x) > 0 \) implies that the function is concave up for \( x \geq 0 \)

Inflection points occur where \( f''(x)=0 \), but none exist here in the defined domain. In this context, a consistently concave-up function ensures a smooth upward curve through its domain, providing a steady rate of increase without abrupt changes in direction.