Problem 16 - Single Variable Calculus

Question

Determine where the graph of the function is concave upward and where it is concave downward. Also, find all inflection points of the function. $$ g(x)=2 x-x^{1 / 3} $$

Step-by-Step Solution

Answer

The graph of the function \(g(x) = 2x - x^{\frac{1}{3}}\) is concave upward when \(x > 0\) and concave downward when \(x < 0\). The inflection point is (0, 0).

Step 1: Find the first derivative g'(x)

To find where the function is concave up or down and the inflection points, we need to first find the first derivative of g(x). The function g(x) is given by: $$ g(x) = 2x - x^{\frac{1}{3}} $$ Differentiate g(x) with respect to x: $$ g'(x) = \frac{d}{dx}(2x) - \frac{d}{dx}(x^{\frac{1}{3}}) $$ Applying the power rule, we have: $$ g'(x) = 2 - \frac{1}{3}\cdot x^{-\frac{2}{3}} $$

Step 2: Find the second derivative g''(x)

Now we need to find the second derivative by differentiating the first derivative g'(x) with respect to x: $$ g''(x) = \frac{d}{dx}(2) - \frac{d}{dx}\left(\frac{1}{3}\cdot x^{-\frac{2}{3}}\right) $$ Applying the power rule again, we get: $$ g''(x) = 0 + \frac{2}{9}x^{-\frac{5}{3}} $$

Step 3: Determine concavity and inflection points

Now we want to know where the second derivative is positive (concave upward) or negative (concave downward), and if the second derivative changes sign, we would find the inflection points. Let's analyze the second derivative: $$ g''(x) = \frac{2}{9}x^{-\frac{5}{3}} $$ Observe that g''(x) > 0 if x > 0 and g''(x) < 0 if x < 0. Thus, the function g(x) is concave upward when x > 0 and concave downward when x < 0. Next, we need to find the inflection points, which occur when the second derivative changes sign. The second derivative changes sign at \(x = 0\). Let's find the inflection point (x, g(x)): $$ g(0) = 2\cdot 0 - (0)^{\frac{1}{3}} = 0 $$ So, the inflection point is (0, 0).

Step 4: Conclusion

The graph of the function g(x) = 2x - x^{\frac{1}{3}} is: - Concave upward when \(x > 0\) - Concave downward when \(x < 0\) The inflection point is (0, 0).

Key Concepts

First Derivative

The first derivative is a key concept in calculus. It measures the rate at which a function is changing at any given point. For the function \( g(x) = 2x - x^{\frac{1}{3}} \), we want to find the first derivative, \( g'(x) \).

• Start by differentiating each term of the function separately.
• For the term \( 2x \), applying the power rule yields \( \frac{d}{dx}(2x) = 2 \).
• For \( x^{\frac{1}{3}} \), use the power rule: differentiate to get \( \frac{1}{3}x^{-\frac{2}{3}} \).

These steps give us the first derivative:\[g'(x) = 2 - \frac{1}{3}x^{-\frac{2}{3}}\]The first derivative helps determine the slope of the tangent lines to the graph at different points, which in turn informs us about the increasing or decreasing nature of the function.

Second Derivative

The second derivative provides insights into the concavity of a function's graph. It is found by differentiating the first derivative. For \( g'(x) = 2 - \frac{1}{3}x^{-\frac{2}{3}} \), let's find the second derivative, \( g''(x) \).

• Differentiate \( 2 \), which results in \( 0 \), since constants become zero.
• Apply the power rule to \( -\frac{1}{3}x^{-\frac{2}{3}} \): differentiating gives \( \frac{2}{9}x^{-\frac{5}{3}} \).

Thus, the second derivative is:\[g''(x) = \frac{2}{9}x^{-\frac{5}{3}}\]The sign of the second derivative tells us about the graph's concavity:- If \( g''(x) > 0 \), the graph is concave upward.- If \( g''(x) < 0 \), the graph is concave downward.

Inflection Points

Inflection points occur where the graph changes its concavity. At such points, the second derivative changes its sign. We already found:\[g''(x) = \frac{2}{9}x^{-\frac{5}{3}}\]To determine inflection points:

• Set the second derivative to correlate with changes in sign: positive for \( x > 0 \) and negative for \( x < 0 \).

The change in sign at \( x = 0 \) indicates an inflection point. Substituting \( x = 0 \) back into the original function gives:\[g(0) = 2 \cdot 0 - (0)^{\frac{1}{3}} = 0\]Thus, the inflection point is at \((0, 0)\). It marks a transition between concave upward and concave downward.

Power Rule

The power rule in calculus is a straightforward technique especially useful for differentiating functions of the form \( x^n \). It states: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).This rule significantly simplifies finding derivatives. Let's look at its application:

• For a constant power, such as \( x^{\frac{1}{3}} \), differentiate to \( \frac{1}{3}x^{-\frac{2}{3}} \).
• Solidly equips us to differentiate terms like \( x^1 \), resulting in \( 1 \cdot x^{1-1} = 1 \).

The power rule makes the process of differentiation much simpler and eliminates complicated computations. It is foundational in calculus and essential for solving many problems, including determining function behavior through derivatives.