Problem 41 - Essential Calculus Early Transcendentals

Question

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(x)=3 x^{4}-4 x^{3}-12 x^{2}+1, \quad[-2,3]$$

Step-by-Step Solution

Answer

Maximum: 77 at x = -2; Minimum: -31 at x = 2.

Step 1: Find the derivative of f(x)

To find the critical points, we first need the derivative of the function. Compute the derivative of \( f(x) = 3x^4 - 4x^3 - 12x^2 + 1 \).\[ f'(x) = \frac{d}{dx}(3x^4) - \frac{d}{dx}(4x^3) - \frac{d}{dx}(12x^2) + \frac{d}{dx}(1) \]\[ f'(x) = 12x^3 - 12x^2 - 24x \]

Step 2: Solve for critical points

Set the derivative equal to zero to find the critical points.\[ 12x^3 - 12x^2 - 24x = 0 \]Factor out the greatest common factor, which is 12x:\[ 12x(x^2 - x - 2) = 0 \]Set each factor to zero:1. \( 12x = 0 \) gives \( x = 0 \).2. \( x^2 - x - 2 = 0 \) can be factored as \( (x - 2)(x + 1) = 0 \), giving \( x = 2 \) and \( x = -1 \).Thus, the critical points are \( x = 0, -1, 2 \).

Step 3: Evaluate f(x) at critical points and endpoints

Calculate \( f(x) \) at each critical point and both endpoints of the interval \([-2, 3]\).At endpoints and critical points:- \( f(-2) = 3(-2)^4 - 4(-2)^3 - 12(-2)^2 + 1 = 77 \)- \( f(-1) = 3(-1)^4 - 4(-1)^3 - 12(-1)^2 + 1 = -8 \)- \( f(0) = 3(0)^4 - 4(0)^3 - 12(0)^2 + 1 = 1 \)- \( f(2) = 3(2)^4 - 4(2)^3 - 12(2)^2 + 1 = -31 \)- \( f(3) = 3(3)^4 - 4(3)^3 - 12(3)^2 + 1 = 28 \)

Step 4: Determine the absolute maximum and minimum values

The absolute maximum occurs at the highest value of \( f(x) \) computed, and the absolute minimum occurs at the lowest value. From Step 3, we calculated:- \( f(-2) = 77 \) - \( f(-1) = -8 \) - \( f(0) = 1 \) - \( f(2) = -31 \) - \( f(3) = 28 \)The absolute maximum value is 77 at \( x = -2 \), and the absolute minimum value is -31 at \( x = 2 \).

Key Concepts

Derivative of a Polynomial

The derivative of a polynomial function is obtained by applying the power rule to every term in the polynomial. Let's take the function \( f(x) = 3x^4 - 4x^3 - 12x^2 + 1 \) as an example. The process involves finding the derivative of each term individually and then combining them. Here are the steps:

• For the term \( 3x^4 \), the derivative is found by multiplying the coefficient (3) by the power of \( x \) (4) and then subtracting one from the power: \( 3 \times 4x^{4-1} = 12x^3 \).
• For \( -4x^3 \), apply the same rule: \( -4 \times 3x^{3-1} = -12x^2 \).
• For \( -12x^2 \), it becomes \( -12 \times 2x^{2-1} = -24x \).
• The term \( +1 \) is constant, and its derivative is zero.

Collectively, the full derivative of \( f(x) \) is \( f'(x) = 12x^3 - 12x^2 - 24x \). This process is vital as it helps locate the critical points of a function where potential maximum or minimum values can occur.

Critical Points

Critical points occur where the derivative \( f'(x) \) equals zero or is undefined. For polynomials like \( f(x) = 3x^4 - 4x^3 - 12x^2 + 1 \), which are smooth and continuous everywhere, we only need to find where \( f'(x) = 0 \). Given:\[ f'(x) = 12x^3 - 12x^2 - 24x = 0 \]you need to factor the expression. Start by extracting the greatest common factor:\[ 12x(x^2 - x - 2) = 0 \].Next, the quadratic \( x^2 - x - 2 \) is factored further:\[ (x - 2)(x + 1) = 0 \].Setting each factor to zero, we solve:

• \( 12x = 0 \) gives \( x = 0 \).
• \( x - 2 = 0 \) results in \( x = 2 \).
• \( x + 1 = 0 \) leads to \( x = -1 \).

Thus, the critical points are \( x = 0 \), \( x = -1 \), and \( x = 2 \). Identifying these points is crucial before evaluating the function to determine maximum or minimum values.

Evaluating Functions at Endpoints

To find the absolute maximum and minimum values of a function within a given interval, evaluate the function at both critical points and endpoints of the interval. This ensures that all possible peaks and troughs within the range are considered. For the polynomial \( f(x) = 3x^4 - 4x^3 - 12x^2 + 1 \) over the interval \([-2, 3]\), calculate:

• \( f(-2) = 3(-2)^4 - 4(-2)^3 - 12(-2)^2 + 1 = 77 \).
• \( f(-1) = 3(-1)^4 - 4(-1)^3 - 12(-1)^2 + 1 = -8 \).
• \( f(0) = 3(0)^4 - 4(0)^3 - 12(0)^2 + 1 = 1 \).
• \( f(2) = 3(2)^4 - 4(2)^3 - 12(2)^2 + 1 = -31 \).
• \( f(3) = 3(3)^4 - 4(3)^3 - 12(3)^2 + 1 = 28 \).

Among these, the absolute maximum value is 77 at \( x = -2 \), and the absolute minimum value is -31 at \( x = 2 \). Evaluating this way ensures that the function's behavior on both ends of the interval is fully accounted for.