Problem 10 - Calculus - AP Edition

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=2 x^{3}+3 x^{2}-12 x ;[-3,2]$$

Step-by-Step Solution

Answer

Maximum value is 20 at \(x = -2\); minimum value is -7 at \(x = 1\).

Step 1: Find the derivative of the function

To find the critical points of the function, we need to take the derivative of the function. Given the function \(f(x) = 2x^3 + 3x^2 - 12x\), we differentiate to get \(f'(x)\).\[f'(x) = \frac{d}{dx}(2x^3 + 3x^2 - 12x) = 6x^2 + 6x - 12.\]

Step 2: Set the derivative equal to zero

To find the critical points, set the derivative equal to zero and solve for \(x\):\[6x^2 + 6x - 12 = 0.\]

Step 3: Solve the quadratic equation

You can solve the equation \(6x^2 + 6x - 12 = 0\) by factoring or using the quadratic formula. First, divide the entire equation by 6 to simplify:\[x^2 + x - 2 = 0.\]This can be factored as:\[(x+2)(x-1) = 0.\]Thus, \(x = -2\) or \(x = 1\) are critical points.

Step 4: Evaluate the function at critical points and endpoints

We need to evaluate the function \(f(x)\) at the critical points and at the endpoints of the interval \([-3, 2]\).Evaluate at \(x = -3\):\[f(-3) = 2(-3)^3 + 3(-3)^2 - 12(-3) = -54 + 27 + 36 = 9.\]Evaluate at \(x = -2\):\[f(-2) = 2(-2)^3 + 3(-2)^2 - 12(-2) = -16 + 12 + 24 = 20.\]Evaluate at \(x = 1\):\[f(1) = 2(1)^3 + 3(1)^2 - 12(1) = 2 + 3 - 12 = -7.\]Evaluate at \(x = 2\):\[f(2) = 2(2)^3 + 3(2)^2 - 12(2) = 16 + 12 - 24 = 4.\]

Step 5: Identify the absolute maximum and minimum values

Compare the values calculated at the critical points and the endpoints: - \(f(-3) = 9\)- \(f(-2) = 20\)- \(f(1) = -7\)- \(f(2) = 4\)The absolute maximum value is 20 at \(x = -2\) and the absolute minimum value is -7 at \(x = 1\).

Key Concepts

Critical Points

To understand critical points, we must first examine where a function's derivative equals zero or is undefined.
In our exercise, the function is a polynomial, and its derivative is given as \(f'(x) = 6x^2 + 6x - 12\).
Critical points are necessary to find local maxima and minima within the interval. Critical points occur at values of \(x\) that make the derivative \(f'(x)\) equal zero.

• First, set \(f'(x)\) to zero: \(6x^2 + 6x - 12 = 0\).
• Here, we solve this quadratic equation to find the critical points.

These critical points help us understand at which \(x\) values the slope of the tangent to the curve is zero, indicating a potential peak or valley.

Absolute Maximum and Minimum

Locating the absolute maximum and minimum values of a function involves determining the highest and lowest points within a particular interval.
In our exercise, we evaluate the function \(f(x) = 2x^3 + 3x^2 - 12x\) over the closed interval \([-3, 2]\).To find these values, examine the function at critical points and boundaries:

• Calculate \(f(x)\) at the critical points \(x = -2\) and \(x = 1\).
• Evaluate \(f(x)\) at the interval endpoints \(x = -3\) and \(x = 2\).

The absolute maximum occurs at the highest function value found: \(f(-2) = 20\), while the absolute minimum is at the lowest value: \(f(1) = -7\). Finding these extremes allows us to discern the complete range covered by the function within the interval.

Derivative

The derivative represents the rate of change or slope of a function at a given point.
It provides critical insight into the behavior of a function, as seen in our exercise.To derive \(f(x) = 2x^3 + 3x^2 - 12x\):

• Apply the power rule: Differentiate each term separately.
• This gives \(f'(x) = 6x^2 + 6x - 12\).

By setting \(f'(x)\) equal to zero, we determine where the function's slope is zero, marking potential critical points.
Additionally, the derivative helps us justify where the function increases or decreases on the interval.

Quadratic Equation

Quadratic equations are polynomial functions of degree two and are often solved to find roots or critical points. In the context of derivatives, solving a quadratic may pinpoint where the slope of a curve reaches zero.For \(6x^2 + 6x - 12 = 0\):

• Simplify by dividing through by 6 to reduce the equation to \(x^2 + x - 2 = 0\).
• This is a standard quadratic equation, which can be solved by factoring.
• Factoring gives \((x+2)(x-1) = 0\), indicating roots at \(x = -2\) and \(x = 1\).

These roots are critical points at which we evaluate the original function to find maximum and minimum values, underscoring the importance of solving such equations effectively.