Problem 7 - Calculus - AP Edition

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=4 x^{2}-12 x+10 ;[1,2]$$

Step-by-Step Solution

Answer

The absolute maximum is 2 at \(x = 1\) and \(x = 2\); the absolute minimum is 1 at \(x = \frac{3}{2}\).

Step 1: Find the Derivative

To find the critical points of the function on the interval, we first need to find the derivative of the function. The function is given as \[ f(x) = 4x^2 - 12x + 10. \]The derivative of this function with respect to \(x\) is \[ f'(x) = 8x - 12. \]

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points:\[ 8x - 12 = 0. \]Solve for \(x\):\[ 8x = 12 \]\[ x = \frac{12}{8} = \frac{3}{2}. \]Thus, the critical point is \(x = \frac{3}{2}\).

Step 3: Evaluate Function at Endpoints and Critical Points

We now evaluate the function \(f(x)\) at the endpoints of the interval \([1, 2]\) and at the critical point \(x = \frac{3}{2}\).- At \(x = 1\), \[ f(1) = 4(1)^2 - 12(1) + 10 = 4 - 12 + 10 = 2. \]- At \(x = 2\), \[ f(2) = 4(2)^2 - 12(2) + 10 = 16 - 24 + 10 = 2. \]- At \(x = \frac{3}{2}\), \[ f\left(\frac{3}{2}\right) = 4\left(\frac{3}{2}\right)^2 - 12\left(\frac{3}{2}\right) + 10 = 4\times\frac{9}{4} - 18 + 10 = 9 - 18 + 10 = 1. \]

Step 4: Determine the Absolute Maximum and Minimum

Comparing the values from the evaluations:- \(f(1) = 2\)- \(f(2) = 2\)- \(f\left(\frac{3}{2}\right) = 1\)The absolute maximum value is 2, which occurs at both endpoints \(x = 1\) and \(x = 2\). The absolute minimum value is 1, which occurs at \(x = \frac{3}{2}\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that allow us to understand the rate at which a function changes at any given point. In simpler terms, the derivative helps us find how steep a curve is at a particular point, which is represented as the slope of the tangent line at that point on a graph. To compute the derivative of a function, we use rules like the power rule, product rule, and chain rule.

For the function given in the exercise, which is a quadratic function of the form \( f(x) = 4x^2 - 12x + 10 \), finding the derivative involves applying the power rule. Using the rule, the derivative is: \[ f'(x) = 8x - 12. \] By taking this derivative, we are equipped to find critical points, which are essential to identify the maximum and minimum values on the interval.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are crucial as they can indicate where a function might have local maxima or minima.

To find these critical points for the given function, we set its derivative \( f'(x) = 8x - 12 \) equal to zero and solve for \( x \). This yields:

• \( 8x - 12 = 0 \)
• \( 8x = 12 \)
• \( x = \frac{3}{2}. \)

This critical point, \( x = \frac{3}{2} \), lies within the interval \([1, 2]\) and becomes a candidate for checking whether it leads to a local maximum or minimum when evaluated in context with the endpoints of the interval.

Quadratic Functions

Quadratic functions are polynomial functions of degree two and can be generally represented as \( ax^2 + bx + c \). These functions graph as parabolas and have vital properties such as symmetry about their vertex. The vertex is the highest or lowest point of the parabola, depending on the direction it opens.

In the given function \( f(x) = 4x^2 - 12x + 10 \), the coefficients \( a = 4 \), \( b = -12 \), and \( c = 10 \), define it as a quadratic function that opens upwards since \( a > 0 \). This upward-opening parabola means any critical point located inside the interval will be a minimum. To find the absolute extreme values over an interval, as described, requires evaluating this quadratic function at:

• the endpoints - \( x = 1 \) and \( x = 2 \)
• the critical point - \( x = \frac{3}{2} \)

Upon performing these evaluations, we confirm the absolute minimum value occurs at the critical point, while the maximums occur at both endpoints.