Problem 16 - Applied Calculus

Question

Find all critical points and then use the first-derivative test to determine local maxima and minima. Check your answer by graphing. $$f(x)=3 x^{4}-4 x^{3}+6$$

Step-by-Step Solution

Answer

Critical points at x=0 (neither) and x=1 (local min, f(1)=5).

Step 1: Find critical points

\(f'(x) = 12x^3 - 12x^2 = 12x^2(x-1) = 0\)

Critical points: \(x = 0\) and \(x = 1\).

Step 2: Apply the first-derivative test

Sign analysis of \(f'(x) = 12x^2(x-1)\):

• \(x < 0\): \(12x^2 > 0, (x-1) < 0 \Rightarrow f' < 0\)
• \(0 < x < 1\): \(12x^2 > 0, (x-1) < 0 \Rightarrow f' < 0\)
• \(x > 1\): \(12x^2 > 0, (x-1) > 0 \Rightarrow f' > 0\)

Step 3: Classify the critical points

At \(x = 0\): \(f'\) does not change sign (negative on both sides), so \(x = 0\) is \(\textbf{neither a max nor a min}\).

At \(x = 1\): \(f'\) changes from negative to positive, so \(x = 1\) is a \(\textbf{local minimum}\).
\(f(1) = 3(1) - 4(1) + 6 = 5\).

Key Concepts

First-Derivative Test

When we want to determine where a function has local maxima or minima, the first-derivative test is a great tool to use. This method involves using the calculated derivative of a given function, in this case, \( f'(x) = 12x^3 - 12x^2 \), to look for critical points and analyze their nature. Critical points are the values of \( x \) where the derivative \( f'(x) \) equals zero or does not exist. To find them, we set the derivative equal to zero:\[12x^3 - 12x^2 = 0\]To solve this equation, we can factor out the greatest common factor:\[12x^2(x-1) = 0\]This gives us \( x = 0 \) and \( x = 1 \) as critical points. Once we have these critical points, we apply the first-derivative test by checking the sign of the derivative before and after each critical point:

• If \( f'(x) \) changes from positive to negative, \( f(x) \) has a local maximum at that point.
• If \( f'(x) \) changes from negative to positive, \( f(x) \) has a local minimum at that point.
• If \( f'(x) \) does not change sign, there might be no extremum at that point.

Perform the test around \( x = 0 \) and \( x = 1 \) to clearly determine their nature.

Local Maxima and Minima

Local maxima and minima are the "peaks" and "valleys" in the graph of a function. They are the points where a function reaches its highest or lowest point within a small vicinity. Using the first-derivative test, we established critical points for our function \( f(x) \), specifically \( x = 0 \) and \( x = 1 \). Now by evaluating these critical points, we can determine the nature of each:

• For \( x = 0 \), if the derivative changes from negative to positive, it signifies a local minimum.
• For \( x = 1 \), if the derivative changes from positive to negative, it indicates a local maximum.

The significance of determining these points is seen not only in calculus but also in various real-world applications. For instance, in economics, local maxima can represent maximum profit points, and local minima might denote minimal cost points. Understanding and identifying these points are key in optimizing any function's outcome in practical scenarios.

Graphing Functions

Graphing functions is a powerful visual tool that helps us understand the behavior of functions, including where they increase, decrease, and the location of critical and inflection points. Graphs provide a clear image of the function's overall trend and help verify algebraic solutions.After finding and testing the critical points of \( f(x) = 3x^4 - 4x^3 + 6 \), graphing the function can confirm the results obtained through the first-derivative test. To graph, plot the function showing:

• The curve of the polynomial function to see its general shape.
• The critical points at \( x = 0 \) and \( x = 1 \) and observe whether they conform to being maxima or minima.
• The intervals where the function is increasing or decreasing around each critical point.

By drawing the function, we can visually verify that \( f(x) \) has a local minimum at \( x = 0 \) and a local maximum at \( x = 1 \). Furthermore, graphing these points reinforces the analysis obtained using calculus, making it a robust approach for understanding complex functions.