Problem 32 - Essential Calculus Early Transcendentals

Question

(a) Find the intervals of increase or decrease. (b) Find the local maximum and minimum values. (c) Find the intervals of concavity and the inflection points. (d) Use the inforvation from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one. \(G(x)=5 x^{2 / 3}-2 x^{5 / 3}\)

Step-by-Step Solution

Answer

(a) Increase: \((-\infty, 1)\); Decrease: \((1, \infty)\). (b) Local max at \( (1, 3) \). (c) Concave down on \((0, \infty)\), no inflection points.

Step 1: Find the Derivative

To determine the intervals of increase or decrease, we first need to find the derivative of the function. Given \[ G(x) = 5x^{2/3} - 2x^{5/3}, \]the derivative is found using the power rule:\[ G'(x) = \frac{d}{dx}(5x^{2/3}) - \frac{d}{dx}(2x^{5/3}) = 5 \cdot \frac{2}{3} x^{-1/3} - 2 \cdot \frac{5}{3} x^{2/3}. \]Simplifying, we get:\[ G'(x) = \frac{10}{3} x^{-1/3} - \frac{10}{3} x^{2/3}. \]Simplify further:\[ G'(x) = \frac{10}{3} \left(x^{-1/3} - x^{2/3}\right). \]

Step 2: Determine Intervals of Increase/Decrease

Set the derivative equal to zero to find critical points:\[ \frac{10}{3} \left(x^{-1/3} - x^{2/3}\right) = 0. \]Simplifying gives:\[ x^{-1/3} = x^{2/3}. \]This can be expressed as:\[ \frac{1}{x^{1/3}} = x^{2/3}. \]Rewriting gives:\[ 1 = x. \]So the critical point is at \(x = 1\). Next, test intervals \((-\infty, 1)\) and \((1, \infty)\) using the derivative:- For \(x < 1\), pick \(x = 0\): \[ G'(0) = \frac{10}{3} \left(0^{-1/3} - 0^{2/3}\right) = \infty. \] The sign is positive (increases).- For \(x > 1\), pick \(x = 2\): \[ G'(2) = \frac{10}{3} \left((2)^{-1/3} - (2)^{2/3}\right) < 0. \] The sign is negative (decreases).Thus, the function increases on \((-\infty, 1)\) and decreases on \((1, \infty)\).

Step 3: Identify Local Extrema

From Step 2, we see that \(x = 1\) is a critical point where the function changes from increasing to decreasing, indicating a local maximum. Calculate \(G(1)\):\[ G(1) = 5(1)^{2/3} - 2(1)^{5/3} = 5 - 2 = 3. \]Thus, there is a local maximum at \((1, 3)\).

Step 4: Determine Concavity and Inflection Points

Find the second derivative to determine concavity:\[ G'(x) = \frac{10}{3} \left(x^{-1/3} - x^{2/3}\right) \Rightarrow G''(x) = \frac{10}{3} \left(-\frac{1}{3} x^{-4/3} - \frac{2}{3} x^{-1/3}\right). \]Simplify:\[ G''(x) = -\frac{10}{9} x^{-4/3} - \frac{20}{9} x^{-1/3}. \]Set the second derivative equal to zero to find inflection points:\[ -\frac{10}{9} x^{-4/3} - \frac{20}{9} x^{-1/3} = 0. \]Solving gives no real solutions, hence no inflection points.However, examine the sign of \(G''(x)\) to find intervals of concavity:- For \(x > 0\), \(G''(x) < 0\) indicating concave down.- For \(x < 0\), \(G''(x)\) is not defined due to the fact that both negative powers are undefined at \(x < 0\). Thus, \(G(x)\) is concave down on \((0, \infty)\).

Step 5: Sketch the Graph

Using the information:- The function increases on \((-\infty, 1)\) and decreases on \((1, \infty)\).- There is a local maximum at \((1, 3)\).- The function is concave down on \((0, \infty)\), with no inflection points.Draw the graph with the axis, marking the local maximum point, and properly adjust the curve to show increasing and decreasing intervals as well as the concavity on \((0, \infty)\).

Key Concepts

Derivative

A derivative is a fundamental concept in calculus, capturing the rate at which a function changes. When you find the derivative of a function, you essentially find a new function that tells you the slope of the original function at any given point. This slope is how steep the curve is at that point, and it helps in understanding how the function behaves.

To calculate the derivative of the function \( G(x) = 5x^{2/3} - 2x^{5/3} \), we use the power rule. The power rule states that for a function \( f(x) = x^n \), the derivative \( f'(x) \) is \( nx^{n-1} \). Applying this to each term of \( G(x) \), we find:

• The derivative of \( 5x^{2/3} \) is \( \frac{10}{3}x^{-1/3} \).
• The derivative of \( 2x^{5/3} \) is \( \frac{10}{3}x^{2/3} \).

Subtracting the second derivative from the first, we obtain \( G'(x) = \frac{10}{3} \left(x^{-1/3} - x^{2/3}\right) \).

This derivative helps us analyze where the function is increasing or decreasing. We can find critical points where the slope is zero or undefined, which helps understand the behavior of the original function.

Critical Points

Critical points occur where the function's derivative is zero or undefined. These points are important because they could be locations of local maximums or minimums. To find them, we set the derivative equal to zero.

For our function \( G'(x) = \frac{10}{3} \left(x^{-1/3} - x^{2/3}\right) \), setting \( G'(x) = 0 \) leads to the equation \( x^{-1/3} = x^{2/3} \). Solving this equation, we find that the critical point is at \( x = 1 \).

We test intervals around this critical point to determine whether it's a local maximum or minimum:

• For values less than 1, \( G'(x) \) is positive, indicating the function is increasing.
• For values greater than 1, \( G'(x) \) is negative, indicating the function is decreasing.

Thus, \( x = 1 \) is a point where the function changes from increasing to decreasing. This change indicates a local maximum.

Concavity

The concavity of a function describes how the slope of the function's graph changes. If a function is concave up, its slope is increasing; if it is concave down, its slope is decreasing. Concavity is examined using the second derivative.

For the function \( G(x) = 5x^{2/3} - 2x^{5/3} \), we found the second derivative: \[ G''(x) = -\frac{10}{9} x^{-4/3} - \frac{20}{9} x^{-1/3}. \] Examining \( G''(x) \) tells us about the function's concavity:

• For \( x > 0 \), \( G''(x) < 0 \), indicating that \( G(x) \) is concave down.
• For \( x < 0 \), \( G''(x) \) is undefined due to negative powers, so the analysis focuses on \( x > 0 \).

Knowing the concavity is crucial as it influences the shape of the graph, helping us understand whether sections of the graph curve upwards or downwards.

Inflection Points

Inflection points are where a function changes its concavity, from concave up to concave down or vice versa. Finding these points involves setting the second derivative equal to zero and solving for \( x \). However, not having real solutions in our second derivative means that no change in concavity happens.

In the case of \( G(x) \), the second derivative is \[ G''(x) = -\frac{10}{9} x^{-4/3} - \frac{20}{9} x^{-1/3} \] which, when set to zero, doesn't yield any real \( x \) values. This implies that there are no inflection points for this function.

Without inflection points, the concavity we determined

• (i.e., function is concave down on \( (0, \infty) \))

is consistent throughout the interval, making the graph appear even smoother over these regions.