Problem 23 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f \), if any, on the given interval, and state where those values occur. \(f(x)=4 x^{3}-3 x^{4} ;(-\infty,+\infty)\)

Step-by-Step Solution

Answer

No absolute extrema on \((-\infty, \infty)\); local max at \( x = 1, f(1) = 1 \).

Step 1: Find the critical points

To find the critical points, we need to compute the derivative of the function and set it equal to zero. The function is given as:\[ f(x) = 4x^3 - 3x^4 \]Compute the derivative:\[ f'(x) = 12x^2 - 12x^3 \]Set the derivative equal to zero to find critical points:\[ 12x^2 - 12x^3 = 0 \]Factor out the greatest common factor:\[ 12x^2 (1 - x) = 0 \]This implies:\[ x = 0 \text{ or } x = 1 \] are the critical points.

Step 2: Second derivative test for local extrema

Use the second derivative test to determine the nature of the critical points. Compute the second derivative of the function:\[ f''(x) = 24x - 36x^2 \]Evaluate the second derivative at the critical points:- At \( x = 0 \), \( f''(0) = 0 \): The test is inconclusive.- At \( x = 1 \), \( f''(1) = 24 - 36 = -12 \): Since it's negative, \( x = 1 \) is a local maximum.

Step 3: Analyze end-behavior as x approaches infinity and negative infinity

Analyze the end behavior of the function \( f(x) = 4x^3 - 3x^4 \).- As \( x \to +\infty \), \( f(x) \to -\infty \) because the \(-3x^4\) term dominates.- As \( x \to -\infty \), \( f(x) \to +\infty \) due to the same reason with higher powers flipping sign accordingly.From this, we can see that there's no absolute minimum or maximum over \((-\infty, +\infty)\).

Step 4: Evaluate the function at critical points

Evaluate the function at critical points to identify local extrema values:- At \( x = 0 \): \( f(0) = 4(0)^3 - 3(0)^4 = 0 \)- At \( x = 1 \): \( f(1) = 4(1)^3 - 3(1)^4 = 1 \)Thus, the local maximum value is 1 at \( x = 1 \) and the function value at \( x = 0 \) is 0.

Key Concepts

Absolute Maximum and Minimum

Finding the absolute maximum and minimum involves identifying the highest and lowest values a function reaches over a set interval. For the function given, \( f(x) = 4x^3 - 3x^4 \), our interval is \(( -\infty, +\infty)\).

Here's how to determine these values step by step:

• First, find the derivative and set it to zero to locate critical points in the function.
• Compute the derivative: \( f'(x) = 12x^2 - 12x^3 \).
• Set \( f'(x) = 0 \) giving \( 12x^2(1 - x) = 0 \), leading to critical points \( x = 0 \) and \( x = 1 \).

To find the absolute maximum and minimum, analyze these critical points. Because the interval is infinite, explore whether the function approaches limits:

• Since no absolute extremum exists over \(( -\infty, +\infty)\), we conclude local extrema exist but not absolute.

End Behavior

The end behavior of a function tells us how it behaves as \( x \) moves towards the edges of the interval, particularly toward positive and negative infinity. This is crucial when considering functions with no clear boundaries for maxima or minima. Let's investigate the end behavior of the function \( f(x) = 4x^3 - 3x^4 \).

To determine this:

• As \( x \to +\infty \), the function tends towards \(-\infty\). This occurs because the term \( -3x^4 \) dominates as it becomes larger due to higher power.
• Conversely, as \( x \to -\infty \), the function heads toward \(+\infty\). In this case, the term with higher power still dominates but the sign changes due to even power and the negative factor.

Understanding end behavior helps confirm that there are neither absolute minimums nor maximums over infinite intervals, guiding further analytical approaches.

Second Derivative Test

The second derivative test helps us determine whether a critical point is a local maximum, minimum, or inconclusive. For the function \( f(x) = 4x^3 - 3x^4 \), we first discovered critical points \( x = 0 \) and \( x = 1 \).

Here's how to proceed with the second derivative test:

• Compute the second derivative: \( f''(x) = 24x - 36x^2 \).
• Evaluate this at each critical point:
• At \( x = 0 \), \( f''(0) = 0 \), making this test inconclusive at this point.
• At \( x = 1 \), \( f''(1) = 24 - 36 = -12 \). The negative value indicates that \( x = 1 \) is a local maximum.

This process aids in understanding the shape and behavior of the graph near critical points, ensuring a thorough analysis of function extrema.