Problem 18 - Applied Calculus

Question

For the functions in problems, do the following: (a) Find \(f^{\prime}\) and \(f^{\prime \prime}\). (b) Find the critical points of \(f\). (c) Find any inflection points of \(f\). (d) Evaluate \(f\) at its critical points and at the endpoints of the given interval. Identify local and global maxima and minima of \(f\) in the interval. (e) Graph \(f\). $$ f(x)=x^{3}-3 x^{2} \quad(-1 \leq x \leq 3) $$

Step-by-Step Solution

Answer

Critical points: \(x = 0, 2\). Inflection point: \(x = 1\). Global min: \(-4\) at \(x = -1\). Graph shows behavior. No clear global max.

Step 1: Find the First Derivative

To find the first derivative \( f'(x) \), differentiate the given function \( f(x) = x^3 - 3x^2 \). Using basic differentiation rules: \[ f'(x) = 3x^2 - 6x. \]

Step 2: Find the Second Derivative

Now, differentiate \( f'(x) = 3x^2 - 6x \) to obtain the second derivative: \[ f''(x) = 6x - 6. \]

Step 3: Find the Critical Points

Critical points occur where the first derivative is zero or undefined. Set \( f'(x) = 0 \): \[ 3x^2 - 6x = 0 \] \[ 3x(x-2) = 0 \] \[ x = 0 \] or \( x = 2 \). These are the critical points in the given interval \(-1 \leq x \leq 3\).

Step 4: Find Inflection Points

Inflection points occur where the second derivative changes sign. Set \( f''(x) = 0 \): \[ 6x - 6 = 0 \] \[ x = 1. \] Check the intervals around \( x = 1 \) to confirm a sign change, establishing it as an inflection point.

Step 5: Evaluate f at Critical Points and Endpoints

Calculate \( f(x) \) at the critical points and endpoints: - \( f(-1) = (-1)^3 - 3(-1)^2 = -1 - 3 = -4 \) - \( f(0) = 0^3 - 3(0)^2 = 0 \) - \( f(2) = 2^3 - 3(2)^2 = 8 - 12 = -4 \) - \( f(3) = 3^3 - 3(3)^2 = 27 - 27 = 0 \) Identify maximum and minimum values: local maxima and minima at these points.

Step 6: Identify Local and Global Extrema

Compare the calculated values: - Local Minima at \((-1, -4)\)- Local Maxima at both \((0, 0)\) and \((3, 0)\) - The function has no global maxima, but global minima occur at the local minimum value of \(-4 \).

Step 7: Graph the Function

Graph \( f(x) = x^3 - 3x^2 \) over the interval \([-1, 3]\) accurately representing critical and inflection points. This visualization helps understand the behavior of the function, confirming the analysis.

Key Concepts

Critical Points

Critical points are an essential concept in calculus. These points occur in a function where the first derivative, \( f'(x) \), is zero or undefined. In simple terms, they are where the slope of the tangent to the curve is flat, indicating potential maxima, minima, or points of inflection.

To find critical points, differentiate the function and set this derivative equal to zero. For example, given \( f(x) = x^3 - 3x^2 \), the first derivative is \( 3x^2 - 6x \). Setting \( f'(x) = 0 \), solve for \( x \):

• \( 3x(x-2) = 0 \)
• Thus, \( x = 0 \) or \( x = 2 \)

These points are where the function changes direction and are critical for understanding the function's behavior between defined intervals. In the interval \([-1, 3]\), \( x = 0 \) and \( x = 2 \) are crucial for further analysis.

Inflection Points

Inflection points represent locations on the curve where the concavity changes. A function changes from concave up (shaped like a cup) to concave down (shaped like a cap), or vice versa, at these points.

To determine inflection points, find where the second derivative, \( f''(x) \), equals zero or changes sign. For our example function, \( f''(x) = 6x - 6 \). Setting \( f''(x) = 0 \) leads us to:

• \( 6x - 6 = 0 \)
• So, \( x = 1 \)

Next, check the sign of \( f''(x) \) on either side of \( x = 1 \):

• If \( x < 1 \), \( f''(x) < 0 \) (concave down)
• If \( x > 1 \), \( f''(x) > 0 \) (concave up)

This confirms \( x = 1 \) as an inflection point, showing the point where the curvature of the graph changes.

Local and Global Extrema

Extrema refers to the highest and lowest points in a specific part of a function's graph. Understanding local and global extrema helps to identify peaks and troughs of a function.

**Local Extrema:** These are the highest or lowest points within a small range of the function. The function \( f(x) = x^3 - 3x^2 \) has critical points at \( x = 0 \) and \( x = 2 \), which can be examined for local maxima or minima.

**Global Extrema:** These represent the absolute highest and lowest values over the function's entire domain. It is crucial to evaluate the function not only at critical points but also at the boundaries of the domain, \(-1\) and \(3\):

• At \( x = -1 \), \( f(-1) = -4 \)
• At \( x = 0 \), \( f(0) = 0 \)
• At \( x = 2 \), \( f(2) = -4 \)
• At \( x = 3 \), \( f(3) = 0 \)

In this case, local minima occur at \((-1, -4)\) and \((2, -4)\), while local maxima are at both \((0, 0)\) and \((3, 0)\). The global minimum value is \(-4\), as this is the lowest value across the entire interval. There are no global maxima as higher values don’t exist outside the local context in this domain.