Problem 80 - Precalculus Mathematics for Calculus

Question

Maximum of a Fourth-Degree Polynomial Find the maximum value of the function $$f(x)=3+4 x^{2}-x^{4}$$ I Hint: Let \(t=x^{2} .1\)

Step-by-Step Solution

Answer

The maximum value of the function is 7.

Step 1: Substitute and Simplify

Given the function \(f(x) = 3 + 4x^2 - x^4\), substitute \(t = x^2\). Therefore, the function becomes \(f(t) = 3 + 4t - t^2\). We need to find the maximum of the function \(f(t)\).

Step 2: Find the Derivative

Find the derivative of \(f(t)\) with respect to \(t\). The derivative is: \(f'(t) = \frac{d}{dt}(3 + 4t - t^2) = 4 - 2t\).

Step 3: Solve for Critical Points

Set the derivative equal to zero and solve for \(t\): \(4 - 2t = 0\). Solving this, we find \(t = 2\).

Step 4: Determine Maximum or Minimum

Evaluate the second derivative to determine whether \(t = 2\) is a maximum or minimum. The second derivative is \(f''(t) = -2\), which is negative, indicating a maximum at \(t = 2\).

Step 5: Substitute Back to Find Maximum Value

Substitute \(t = 2\) back into the function \(f(t)\) to find the maximum value. \(f(2) = 3 + 4(2) - (2)^2 = 3 + 8 - 4 = 7\). The maximum value of the function is 7.

Key Concepts

Fourth-degree polynomial

A fourth-degree polynomial is a type of mathematical expression that involves a variable raised to the power of four. These expressions can take the form of \( ax^4 + bx^3 + cx^2 + dx + e \), where \( a, b, c, d, \) and \( e \) are constants and \( a eq 0 \).

In the exercise, we are working with the polynomial \( f(x) = 3 + 4x^2 - x^4 \). This particular polynomial does not contain a cubic term (\( x^3 \)), and its leading term \( -x^4 \) indicates that the polynomial will eventually decrease to negative infinity as \( x \) moves away from zero.

• Fourth-degree polynomials can be more complex due to additional curvature compared to lower-degree polynomials.
• They can have up to four roots or zero points where the expression equals zero.
• The importance in calculus comes from being able to identify the highest or lowest points (maximums or minimums) which are often needed in practical applications such as engineering.

Understanding these basic features helps in analyzing their behavior and applying calculus concepts to find key points such as maxima.

Critical points

Critical points are essential elements in calculus used to identify the behavior of a function. They occur where the derivative of a function is zero or undefined.

For the function \( f(t) = 3 + 4t - t^2 \), its derivative \( f'(t) \) becomes zero at \( t = 2 \). This point requires further analysis to determine if it represents a maximum or minimum value of the function.

• Critical points can indicate potential peaks (maxima), valleys (minima), or saddle points in the graph of the function.
• Identifying critical points allows us to explore changes in the function's behavior, such as increasing or decreasing trends.

Recognizing and analyzing critical points is a crucial step toward understanding and optimizing a function.

Derivative

The derivative is a fundamental concept in calculus which represents the rate of change of a function. It provides information about the slope or gradient of the function at any given point.

In this exercise, we derive \( f(t) = 3 + 4t - t^2 \) to obtain \( f'(t) = 4 - 2t \), a simple linear expression. This derivative helps us locate critical points by setting \( f'(t) = 0 \).

• The sign of the derivative tells us if a function is increasing (positive) or decreasing (negative).
• By working out the derivative, we can discover where a function flattens out or changes direction.

Using derivatives, one can solve numerous real-world problems, such as maximizing efficiency or increasing productivity.

Second derivative test

The second derivative test is a method used to classify critical points further, distinguishing between maxima, minima, and points of inflection.

For the function \( f(t) = 3 + 4t - t^2 \), the second derivative \( f''(t) = -2 \) is constant and negative, indicating that the function is concave down everywhere. Hence, the critical point at \( t = 2 \) represents a local maximum.

• If the second derivative is positive at a critical point, the point is a minimum.
• If the second derivative is negative, the point is a maximum.
• A zero second derivative might indicate an inflection point, requiring further analysis.

Using the second derivative test is a powerful technique in calculus to determine the nature of critical points and make more informed decisions based on the function's behavior.