Problem 20 - 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 \(f\). $$ f(x)=2 x^{3}-9 x^{2}+12 x+1(-0.5 \leq x \leq 3) $$

Step-by-Step Solution

Answer

The critical points are \(x = 1\) and \(x = 2\); inflection point at \(x = 1.5\). The maximum is at \(x = 1\) and the minimum at \(x = 3\).

Step 1: Find the First Derivative, \(f'(x)\)

To find the first derivative of \(f(x) = 2x^3 - 9x^2 + 12x + 1\), apply the power rule: - The derivative of \(2x^3\) is \(6x^2\).- The derivative of \(-9x^2\) is \(-18x\).- The derivative of \(12x\) is \(12\).- The derivative of constant \(1\) is \(0\).Therefore, \(f'(x) = 6x^2 - 18x + 12\).

Step 2: Find the Second Derivative, \(f''(x)\)

Differentiate \(f'(x) = 6x^2 - 18x + 12\) to get the second derivative:- The derivative of \(6x^2\) is \(12x\).- The derivative of \(-18x\) is \(-18\).- The derivative of the constant \(12\) is \(0\).Thus, \(f''(x) = 12x - 18\).

Step 3: Find the Critical Points

Critical points occur where the first derivative is zero or undefined. Set \(f'(x) = 6x^2 - 18x + 12 = 0\).Factor the equation: \[6(x^2 - 3x + 2) = 6((x-1)(x-2)) = 0\]Solve for \(x\):- \(x = 1\)- \(x = 2\)These are the critical points.

Step 4: Find Inflection Points

Inflection points occur where the second derivative is zero or changes sign. Set \(f''(x) = 12x - 18 = 0\).Solve for \(x\):- \(12x - 18 = 0\)- \(12x = 18\)- \(x = 1.5\)Thus, \(x = 1.5\) is a potential inflection point.

Step 5: Evaluate \(f(x)\) at Critical Points and Endpoints

Evaluate \(f(x)\) at critical points \(x = 1\), \(x = 2\), and endpoints \(x = -0.5\) and \(x = 3\):- \(f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 1 = 6\)- \(f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 1 = 1\)- \(f(-0.5) = 2(-0.5)^3 - 9(-0.5)^2 + 12(-0.5) + 1 = -5.5\)- \(f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 1 = -8\)

Step 6: Identify Local and Global Extrema

Compare values:- Maximum is at \(x = 1\) with \(f(1) = 6\) (local and global maximum).- Minimum likely at endpoints: \(x = -0.5, f(-0.5) = -5.5\) and \(x = 3, f(3) = -8\) (global minimum at \(x = 3\)).Critical point \(x = 2, f(2) = 1\) is neither max nor min.

Step 7: Sketch the Graph of \(f(x)\)

Plot points: Critical points \((1, 6)\) and \((2, 1)\), endpoints \((-0.5, -5.5)\) and \((3, -8)\), and the inflection point \((1.5, -0.75)\). Plot curve with maximum at \(x = 1\) and minimal value at the interval endpoints and inflection at \(x = 1.5\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, and they represent the rate at which a function is changing at any point. To find the derivative of a function, we typically apply rules like the power rule, product rule, or chain rule, depending on the function's complexity. In the given function,\[ f(x) = 2x^3 - 9x^2 + 12x + 1, \] we use the power rule to determine the first derivative, \( f'(x) = 6x^2 - 18x + 12.\) Each term of the function is differentiated separately. The power rule states that when differentiating \( x^n, \) the derivative is \( nx^{n-1}.\) Calculating derivatives allows us to understand how a function's output changes concerning changes in input around a particular point. This information is crucial when analyzing the graph of the function, predicting its behavior, and finding points of interest like critical and inflection points. Understanding derivatives helps establish where the function is increasing or decreasing, informing us about possible turning points.

Critical Points

Critical points are specific values of \( x \) where the first derivative of the function is zero or undefined. These points can indicate locations where the graph of the function has a peak or a valley—essentially potential local maxima or minima. For the function \[ f(x) = 2x^3 - 9x^2 + 12x + 1, \] we found the critical points by setting the first derivative \( f'(x) = 6x^2 - 18x + 12 \) to zero and solving the equation:

• \(6(x-1)(x-2) = 0\)
• \(x = 1\)
• \(x = 2\)

These calculations suggest that \( x = 1 \) and \( x = 2 \) are critical points. To confirm these are indeed maxima or minima, we further evaluate the function at these points and compare them to other values such as the endpoints of the interval. Analyzing critical points is significant for understanding the structure of the function's graph and identifying important changes in its direction.

Inflection Points

Inflection points are where a function's graph changes concavity. This occurs where the second derivative is either zero or undefined, indicating a shift from concave up to concave down or vice versa. For our function \( f(x) = 2x^3 - 9x^2 + 12x + 1, \) we find the second derivative \( f''(x) = 12x - 18.\) Setting \( f''(x) = 0, \) we solve:

• \(12x - 18 = 0\)
• \(12x = 18\)
• \(x = 1.5\)

The solution \( x = 1.5 \) is the inflection point, where the graph of the function changes its curvature. This is a crucial aspect in graphing as it identifies where concavity and convexity alternate, influencing the overall shape of the function. These points help provide a comprehensive view of the function’s behavior and guide us in sketching an accurate graph.

Graphing Functions

Graphing functions is essential for visualizing their behavior over a specific interval. To graph the function \( f(x) = 2x^3 - 9x^2 + 12x + 1 \), we incorporate crucial characteristics indicated by derivatives, such as critical points, inflection points, and endpoint evaluations. This function behaves within the interval \(-0.5 \leq x \leq 3.\) Here are the steps:

• Identify critical points at \(x = 1\) and \(x = 2\).
• Determine inflection point at \(x = 1.5\).
• Evaluate the function at important points:
  • \(f(1) = 6\)
  • \(f(2) = 1\)
  • \(f(-0.5) = -5.5\)
  • \(f(3) = -8\)

This function demonstrates a local and global maximum at \(x = 1\) and a minimum value at the endpoints and other critical points. Including all these elements on the graph gives a complete picture of how the function behaves across the given domain, showing where it rises and falls and how its curvature shifts. A precise graph highlights the function's crucial features, supporting better comprehension and interpretation of the mathematical relationship.