Problem 17 - 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 \(f\). $$f(x)=2 x^{3}-9 x^{2}+12 x+1(-0.5 \leq x \leq 3)$$

Step-by-Step Solution

Answer

The local max is at \( x = -0.5 \), and the local min is at \( x = 2 \).

Step 1: Differentiate once to find f'

The function given is \( f(x) = 2x^3 - 9x^2 + 12x + 1 \). To find the first derivative \( f'(x) \), apply the power rule. This gives \( f'(x) = 6x^2 - 18x + 12 \).

Step 2: Differentiate again to find f''

To find the second derivative \( f''(x) \), differentiate \( f'(x) = 6x^2 - 18x + 12 \). This results in \( f''(x) = 12x - 18 \).

Step 3: Find Critical Points

Critical points occur when \( f'(x) = 0 \). Solve \( 6x^2 - 18x + 12 = 0 \) using the quadratic formula. The solutions are \( x = 1 \) and \( x = 2 \).

Step 4: Find Inflection Points

Inflection points occur where \( f''(x) = 0 \). Solve \( 12x - 18 = 0 \) which results in \( x = 1.5 \) as the inflection point.

Step 5: Evaluate at Critical Points and Endpoints

Evaluate \( f(x) \) at critical points and endpoints: \( x = -0.5, 1, 2, 3 \). This gives \( f(-0.5) = 6.125 \), \( f(1) = 6 \), \( f(2) = 1 \), \( f(3) = 4 \).

Step 6: Identify Extrema

From the evaluations, the local maximum is at \( x = -0.5 \) and the local minimum is at \( x = 2 \). There is no global maximum or minimum as the interval is not closed by the graph analysis.

Step 7: Graph the Function

Graph the function using technology or sketch by applying known points including \( x = -0.5, 1, 2, 3 \). Note shape changes at the inflection point \( x = 1.5 \).

Key Concepts

Calculus

Calculus is a branch of mathematics focusing on limits, functions, derivatives, integrals, and infinite series. It deals primarily with change and motion.
The key components in calculus are:

• Limits: These provide a way of understanding the behavior of functions as input approaches a certain point.
• Derivatives: They give us the rate at which a function is changing at any given moment. Calculus uses derivatives to find slopes and rates of change.
• Integrals: Integration is the process of accumulating quantities over time or space, used to find areas under curves.

Calculus is essential for understanding and solving real-world problems in physics, engineering, economics, and more.
It provides tools for modeling and predicting change, helping us analyze scenarios involving dynamic systems.

Inflection Points

Inflection points in calculus are points on a curve where the curvature changes sign.
This means the curve moves from being convex (curving upwards) to concave (curving downwards), or vice versa.
To find inflection points, we examine the second derivative of a function, since it provides information on the concavity of the curve.

• How to identify: Determine where the second derivative, denoted as \( f''(x) \), changes sign from positive to negative or negative to positive.
  For example, if \( f''(x) \)=0 at \( x=a \), investigate this point as an inflection point.
• Importance: Inflection points help to understand the underlying behavior of the graph.
  They signify changes in the slope's rate of change and provide insights into the function’s geometric features.

Derivatives

Derivatives form the cornerstone of differential calculus and represent the rate of change of a function with respect to a variable.
They tell you how a function changes as its input changes.
How do we find derivatives?

• Basic Rule: The most common technique is the power rule: If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \).
• Applying to multiple terms: When dealing with polynomials, differentiate each term according to the power rule.For example, for \( f(x) = 2x^3 - 9x^2 + 12x + 1 \), differentiate to get \( f'(x) = 6x^2 - 18x + 12 \).
• Significance: Derivatives can determine critical points, where a function's slope is zero, indicating potential maxima, minima, or points of inflection.

Local Extrema

Local extrema refer to the local maximum and minimum values of a function in a specific interval.
These are the 'highest' or 'lowest' points in the neighborhood of a graph.

• Local Maximum: Occurs when a function changes from increasing to decreasing, making the point a peak.
• Local Minimum: Happens where a function switches from decreasing to increasing, forming a trough.
• Finding Local Extrema: To identify these points, look for where the first derivative \( f'(x) = 0 \) and confirm with the second derivative sign test.Positive indicates a minimum, while negative indicates a maximum.
• Applications: Local extrema are crucial in optimization problems, helping identify optimal values for functions within specific ranges.

Graphing Functions

Graphing functions is a vital skill in calculus, allowing visualization of function behavior.
Graphs illustrate how a function behaves over an interval and highlight critical points, extrema, and inflection points.

• Key Steps in Graphing:
  • Calculate critical and inflection points using derivatives.
  • Evaluate the function at these points and selected endpoints to get specific graph coordinates.
  • Sketch the curve based on these points and the function's behavior. Consider concavity and sign changes, especially around inflection points.
• Technology in Graphing: Use graphing calculators or software to plot complex functions easily and verify calculations.
• Benefits of Graphing: Graphs offer a clear picture of function trends, making it easier to determine overall behavior and identify key features.