Problem 36 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=x^{2 / 3}(x-5)$$

Step-by-Step Solution

Answer

Local extreme is around \((2, -4.7622)\); inflection points need numerical methods.

Step 1: Find the First Derivative

To find the local extreme points, we first need the first derivative of the function. Given function: \[ y = x^{2/3}(x-5) \] First, let's apply the product rule, which states that \((uv)' = u'v + uv'\).Let \( u = x^{2/3} \) and \( v = (x-5) \).The derivative of \( u \) is:\[ u' = \frac{d}{dx}(x^{2/3}) = \frac{2}{3}x^{-1/3} \]The derivative of \( v \) is:\[ v' = \frac{d}{dx}(x-5) = 1 \]Now apply the product rule:\[ y' = u'v + uv' = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \cdot 1 \]Simplify:\[ y' = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \]

Step 2: Set First Derivative to Zero

To find the critical points, set the first derivative equal to zero:\[ 0 = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \]Combine like terms in a common denominator to simplify:\[ 0 = \frac{2(x-5) + 3x}{3x^{1/3}} \]\[ 0 = \frac{2x - 10 + 3x}{3x^{1/3}} \]\[ 0 = \frac{5x - 10}{3x^{1/3}} \]Solving for \( x \), set the numerator equal to zero:\[ 5x - 10 = 0 \]\[ 5x = 10 \]\[ x = 2 \]

Step 3: Analyze Critical Points

With \( x = 2 \) as a critical point, substitute back into the original function to find the y-coordinate:\[ y = 2^{2/3}(2-5) \]Calculate \( 2^{2/3} \) and \( (2-5) = -3 \):\[ y = 2^{2/3} imes (-3) \]\[ y \approx -3 imes 1.5874 = -4.7622 \]Thus, the coordinates of the local extreme point are approximately \((2, -4.7622)\).

Step 4: Find the Second Derivative

Now let's find the second derivative to identify inflection points.Starting with:\[ y' = \frac{5x - 10}{3x^{1/3}} \]Apply the quotient rule \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\):Let \( u = 5x - 10 \) and \( v = 3x^{1/3} \). Find their derivatives:\[ u' = 5, \quad v' = x^{-2/3} \]Derivative of \( y' \) is:\[ y'' = \frac{(5)(3x^{1/3}) - (5x - 10)(x^{-2/3})}{(3x^{1/3})^2} \]Simplify:\[ y'' = \frac{15x^{1/3} - (5x - 10)x^{-2/3}}{9x^{2/3}} \]\[ y'' = \frac{15x + 10}{9x} - \frac{5}{9x^{1/3}} \]

Step 5: Set Second Derivative to Zero

Set the second derivative to zero to identify potential inflection points:\[ 0 = \frac{15x + 10}{9x} - \frac{5}{9x^{1/3}} \]Finding a common denominator:\[ rac{15x + 10 - 5x^{2/3}}{9x} = 0 \]Set the numerator equal to zero:\[ 15x + 10 - 5x^{2/3} = 0 \]This equation is complex algebraically without a straightforward factorization. For precise inflection points, use numerical methods or graphing.

Step 6: Graph the Function

Graph the function \( y = x^{2/3}(x-5) \) to visualize the identified points.The local extreme point is approximately at \((2, -4.7622)\), visually verified. Inflection points can be estimated graphically where the curvature changes.

Key Concepts

First Derivative

The first derivative of a function, denoted as \( y' \) or \( f'(x) \), gives us crucial information about the function's behavior. It basically tells us the slope of the tangent line to the function at any given point. To find local extreme points like maxima or minima, the first derivative is our best friend. For the function \( y = x^{2/3}(x-5) \), we apply the product rule to derive the first derivative.

• Set \( y = x^{2/3}(x-5) \).
• The first derivative is found using \( u = x^{2/3} \) and \( v = x-5 \).
• Compute \( u' = \frac{2}{3}x^{-1/3} \) and \( v' = 1 \).

Applying the product rule gives us:\[ y' = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \] By setting \( y' = 0 \), we can find the critical points, which are potential candidates for local extreme points.

Critical Points

Critical points occur where the first derivative equals zero or is undefined. Finding these points is essential to understand where a function reaches its highest or lowest values. For our function, we set:
\[ 0 = \frac{2}{3}x^{-1/3}(x-5) + x^{2/3} \]This equation simplifies to:\[ 0 = \frac{5x - 10}{3x^{1/3}} \]
By solving the equation \( 5x - 10 = 0 \), we find \( x = 2 \). This value marks a critical point. To determine if it’s a local minimum or maximum, we need additional analysis like sign testing or using the second derivative.

Second Derivative

The second derivative, denoted as \( y'' \), provides insights into a function's concavity: whether it curves upwards or downwards. It is calculated by differentiating the first derivative. For the second derivative of our function:
\[ y' = \frac{5x - 10}{3x^{1/3}} \]We apply the quotient rule to get:\[ y'' = \frac{15x^{1/3} - (5x - 10)x^{-2/3}}{9x^{2/3}} \]A positive \( y'' \) indicates the graph is concave up, while a negative \( y'' \) implies concave down.
These positive or negative signs help identify whether a critical point is a local minimum or maximum.

Inflection Points

Inflection points are where the curvature of the function changes from up to down or vice versa. This is detected by examining where the second derivative equals zero or does not exist.
Setting:\[ 0 = \frac{15x + 10}{9x} - \frac{5}{9x^{1/3}} \]Solving for zero can be complex, but graphing or numerical methods often aid in finding exact points where curvature changes.
Inflection points signify significant transitions in function behavior, as the graph bends differently at these spots.

Product Rule

The product rule is crucial in differentiating functions that are products of two simpler functions, say \( u(x) \) and \( v(x) \). The rule states:
\[ (uv)' = u'v + uv' \]It applies to our example function, \( y = x^{2/3}(x-5) \), where:

• \( u = x^{2/3}, u' = \frac{2}{3}x^{-1/3} \)
• \( v = x-5, v' = 1 \)

Using the product rule ensures that every aspect of the function's behavior is captured in the derivative, enabling deeper analysis of critical points and shape.

Quotient Rule

When it comes to computing the derivative of a function that presents as a division of two substrings, the quotient rule is used. The quotient rule is given by:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]For the first derivative of our function, the quotient rule helps find the second derivative. It handles functions where one part of the overall function is divided by another, like:

• Numerator: \( 5x - 10 \)
• Denominator: \( 3x^{1/3} \)

Mastering the quotient rule is essential for solving diverse calculus problems efficiently, including those involving complex functions.