Problem 18 - Applied Calculus

Question

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 \(f\). $$f(x)=x^{3}-3 x^{2}-9 x+15 \quad(-5 \leq x \leq 4)$$

Step-by-Step Solution

Answer

Critical points: \( x = -1, 3 \). Inflection point: \( x = 1 \). Global max: \( x = -1 \), global min: \( x=-5 \).

Step 1: Differentiate the Function

First, we need to determine the first and second derivatives of the function \( f(x) = x^3 - 3x^2 - 9x + 15 \).The first derivative \( f'(x) \) is obtained by differentiating each term.\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(9x) + \frac{d}{dx}(15) \]\[ f'(x) = 3x^2 - 6x - 9 \]The second derivative \( f''(x) \) is the derivative of \( f'(x) \).\[ f''(x) = \frac{d}{dx}(3x^2) - \frac{d}{dx}(6x) - \frac{d}{dx}(9) \]\[ f''(x) = 6x - 6 \]

Step 2: Find Critical Points

Critical points occur where \( f'(x) = 0 \) or where \( f'(x) \) is undefined. Since \( f'(x) = 3x^2 - 6x - 9 \), we solve for \( x \) where it equals zero.\[ 3x^2 - 6x - 9 = 0 \]Divide the entire equation by 3:\[ x^2 - 2x - 3 = 0 \]Factor the quadratic equation:\[ (x - 3)(x + 1) = 0 \]Thus, \( x = 3 \) and \( x = -1 \) are critical points.

Step 3: Check for Inflection Points

To find inflection points, set \( f''(x) = 0 \). Inflection points occur where the concavity changes, which is determined by the sign of \( f''(x) \).\[ f''(x) = 6x - 6 = 0 \]Solve for \( x \):\[ 6x = 6 \]\[ x = 1 \]Check the concavity change around \( x = 1 \) to confirm it's an inflection point. Test values on either side of \( x=1 \), confirming the concavity changes, thus it is an inflection point.

Step 4: Evaluate Critical Points and Endpoints

Evaluate \( f(x) \) at the critical points \( x = 3 \), \( x = -1 \), and the endpoints \( x = -5 \) and \( x = 4 \) within the interval \(-5 \leq x \leq 4\).\[ f(-5) = (-5)^3 - 3(-5)^2 - 9(-5) + 15 = -125 - 75 + 45 + 15 = -140 \]\[ f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 15 = -1 - 3 + 9 + 15 = 20 \]\[ f(3) = (3)^3 - 3(3)^2 - 9(3) + 15 = 27 - 27 - 27 + 15 = -12 \]\[ f(4) = (4)^3 - 3(4)^2 - 9(4) + 15 = 64 - 48 - 36 + 15 = -5 \]Identify the extremums: at \( x = -1 \) \( f(x) \) is 20 (local and global maximum), at \( x = 3 \) \( f(x) \) is -12 (local minimum), and at endpoints \( x = -5 \) and \( x = 4 \), \( f(x) \) are -140 and -5, respectively.

Key Concepts

Critical Points

Critical points in calculus are values of \( x \) where the derivative \( f'(x) \) is either zero or undefined. These points are important because they can indicate potential maxima, minima, or saddle points (points which are neither maximum or minimum but still critical in behavior). To identify critical points for a function like \( f(x) = x^3 - 3x^2 - 9x + 15 \), we first find the derivative \( f'(x) = 3x^2 - 6x - 9 \).

• Set this first derivative to zero: \( 3x^2 - 6x - 9 = 0 \).
• Solving this gives the quadratic equation: \( x^2 - 2x - 3 = 0 \).
• Factor this to find \( (x - 3)(x + 1) = 0 \), leading to critical points \( x = 3 \) and \( x = -1 \).

These critical points indicate where the function could have peaks or valleys, essential for determining the overall shape and key features of the curve when graphed.

Inflection Points

Inflection points are places where the concavity of a function changes, going from concave up to concave down, or vice versa. To find them, consider the second derivative \( f''(x) \). For the function \( f(x) = x^3 - 3x^2 - 9x + 15 \), the second derivative is \( f''(x) = 6x - 6 \).

• Set \( f''(x) = 0 \) to find potential inflection points, which yields \( 6x - 6 = 0 \).
• Solve for \( x \) to find the potential inflection point: \( x = 1 \).

To confirm that \( x = 1 \) is truly an inflection point, check the sign of \( f''(x) \) before and after \( x = 1 \):

• Before \( x = 1 \) (say, at \( x = 0 \)), \( f''(x) < 0 \) indicates concavity is down.
• After \( x = 1 \) (say, at \( x = 2 \)), \( f''(x) > 0 \), showing concavity is up.

This change in concavity at \( x = 1 \) confirms it is an inflection point, making the curve shift from a downward bend to an upward bend.

Maxima and Minima

Maxima and minima are the highest and lowest points on a function, respectively, within a given interval. They're crucial for understanding the bounds of a function. For \( f(x) = x^3 - 3x^2 - 9x + 15 \) within the interval \(-5 \leq x \leq 4\), we use the critical points \( x = 3 \) and \( x = -1 \), and also evaluate the function at the interval's endpoints, \( x = -5 \) and \( x = 4 \).

Evaluate the function at these points:

• \( f(-5) = -140 \)
• \( f(-1) = 20 \)
• \( f(3) = -12 \)
• \( f(4) = -5 \)

Comparing these values:

• Maximum (highest value) is at \( x = -1 \) with \( f(x) = 20 \), which is both a local and global maximum within the interval.
• Local minimum occurs at \( x = 3 \) with \( f(x) = -12 \).
• The endpoints help determine overall bounds of the function's behavior but may not be maximums or minimums unless they stand out compared to critical points.

Identifying maximum and minimum points assists in visualizing the function's peaks and troughs, providing an understanding of how the function behaves over the specified interval.