Problem 26 - Calculus Early Transcendentals

Question

More graphing Make a complete graph of the following functions. If an interval is not specified, graph the function on its domain. Use a graphing utility to check your work. $$f(x)=2-x^{2 / 3}+x^{4 / 3}$$

Step-by-Step Solution

Answer

The function is concave up on its entire domain, $$(-\infty, \infty)$$. There are no inflection points, and the function is increasing throughout its domain.

Step 1: Find the domain

The given function is $$f(x)=2-x^{2/3}+x^{4/3}$$, which is a combination of power functions with rational exponents. Since the exponents are non-negative fractions, the function has no restrictions on its domain, and the domain of the function is all real numbers, i.e., $$(-\infty, \infty)$$.

Step 2: Find critical points

To find critical points, we need to find the first derivative of the function and set it equal to zero or be undefined. The first derivative of the function is: $$f'(x)=\frac{d}{dx}\left(2-x^{2/3}+x^{4/3}\right)=-\frac{2}{3}x^{-1/3}+\frac{4}{3}x^{1/3}$$ Now find the critical points by setting the derivative to zero: $$0= -\frac{2}{3}x^{-1/3}+\frac{4}{3}x^{1/3}$$ Moving the negative term to the other side and factoring the common term, we get: $$\frac{2}{3}x^{-1/3}=\frac{4}{3}x^{1/3}$$ $$1=2x^{2/3}$$ $$x^{2/3}=\frac{1}{2}$$ $$x=\sqrt[3]{\frac{1}{4}}$$ The critical point that we found is $$x=\sqrt[3]{\frac{1}{4}}$$.

Step 3: Determine intervals of increasing and decreasing

Now that we have found the critical point, use that to determine the intervals of increasing and decreasing of the function. Make a sign chart of the derivative: 1. On the left side of $$x=\sqrt[3]{\frac{1}{4}}$$, pick a test point, like $$x=0$$, and plug it into the derivative to get $$f'(0)=\frac{4}{3}>0$$, the function is increasing on \((-\infty, \sqrt[3]{\frac{1}{4}})\). 2. On the right side of $$x=\sqrt[3]{\frac{1}{4}}$$, pick a test point, like $$x=1$$, and plug it into the derivative to get $$f'(1)=-\frac{2}{3}+\frac{4}{3}>0$$, the function is increasing on \((\sqrt[3]{\frac{1}{4}}, \infty)\). Now that we know the function is increasing throughout its domain, there is no local minimum or maximum.

Step 4: Find concavity and inflection points

To find the concavity and inflection points, find the second derivative and analyze its sign. The second derivative of the function is: $$f''(x)=\frac{d^2}{dx^2}\left(2-x^{2/3}+x^{4/3}\right) = \frac{2}{9}x^{-4/3}+\frac{4}{9}x^{-2/3}$$ There are no critical points in the second derivative, so we can only analyze the concavity on the entire domain. Choose a test point, like $$x=1$$, and plug it into the second derivative, we get: $$f''(1)=\frac{2}{9}+\frac{4}{9}>0$$ Thus, the function is concave up on \((-\infty, \infty)\). There are no inflection points.

Step 5: Sketch the graph and check with a graphing utility

Now, gather all the information: - Domain: $$(-\infty, \infty)$$ - Critical point: $$x=\sqrt[3]{\frac{1}{4}}$$ - Increasing: $$(-\infty, \infty)$$ - Concave up: $$(-\infty, \infty)$$ - No local extrema or inflection points. Sketch the graph by including the critical point and observing the behavior in terms of increasing and concavity. Your graph should show a smooth curve that starts on the lower left corner, increases as it passes through the critical point, and continues increasing towards the upper right corner. To verify your graph, use a graphing utility such as Desmos, GeoGebra, or a graphing calculator. Input the function $$2-x^{2/3}+x^{4/3}$$ and compare its graph with the sketch. They should be consistent.

Key Concepts

Critical Points

Critical points are where the first derivative of a function is zero or undefined. Finding these points helps us identify where a function might have local maxima or minima. For the function \( f(x) = 2 - x^{2/3} + x^{4/3} \), we calculate the first derivative:

• First derivative, \( f'(x) = -\frac{2}{3} x^{-1/3} + \frac{4}{3} x^{1/3} \)
• Set \( f'(x) = 0 \) to find critical points
• Solving \( 0 = -\frac{2}{3} x^{-1/3} + \frac{4}{3} x^{1/3} \) gives \( x = \sqrt[3]{\frac{1}{4}} \)

So, the critical point is \( x = \sqrt[3]{\frac{1}{4}} \). At this point, the slope of the tangent to the curve is zero, indicating a potential change in the behavior of the function.

Intervals of Increase and Decrease

Determining the intervals where the function is increasing or decreasing involves examining the sign of the first derivative.

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

With \( f'(x) = -\frac{2}{3} x^{-1/3} + \frac{4}{3} x^{1/3} \), test the intervals around \( x = \sqrt[3]{\frac{1}{4}} \):

• For \( x = 0 \), \( f'(0) = \frac{4}{3} > 0 \) indicating the function is increasing in \( (-\infty, \sqrt[3]{\frac{1}{4}}) \).
• For \( x = 1 \), \( f'(1) = -\frac{2}{3} + \frac{4}{3} > 0 \) indicating the function is also increasing in \( (\sqrt[3]{\frac{1}{4}}, \infty) \).

Thus, the function is increasing on both sides of the critical point, suggesting no local maxima or minima within these intervals.

Concavity and Inflection

Concavity tells us about the direction the curve bends, and inflection points are where the concavity changes. We determine concavity by examining the second derivative.

• Second derivative, \( f''(x) = \frac{2}{9}x^{-4/3} + \frac{4}{9}x^{-2/3} \)

Check the sign of \( f''(x) \):

• If \( f''(x) > 0 \), the function is concave up.
• If \( f''(x) < 0 \), the function is concave down.

Calculate \( f''(1) = \frac{2}{9} + \frac{4}{9} > 0 \), showing the function is concave up for all \( x \).
Since the second derivative does not change sign, there are no inflection points, and the concavity remains constant over the domain.

Domain of a Function

The domain of a function is the set of all possible input values (\( x \)-values) for which the function is defined. For the given function \( f(x) = 2 - x^{2/3} + x^{4/3} \), the expression involves rational exponents, but only positive ones.

• This means it does not impose any additional restrictions on \( x \).

Hence, the domain of \( f \) includes all real numbers, or \( (-\infty, \infty) \). This implies you can plug in any real number for \( x \) without worrying about the function being undefined.
Understanding the domain is crucial in determining where the function can be graphed and analyzed. It forms the basis for examining critical points, intervals, concavity, and more.