Problem 15 - Calculus - AP Edition

Question

Find the absolute maximum and minimum values of \(f\) on the given closed interval, and state where those values occur. $$f(x)=1+\left|9-x^{2}\right| ;[-5,1]$$

Step-by-Step Solution

Answer

Maximum: 17 at x = -5; Minimum: 9 at x = 1.

Step 1: Determine Critical Points

Identify the points where the derivative of the function may equal zero or is undefined in order to find potential candidates for extreme values. The function is given by \[f(x) = 1 + |9-x^2|\] We need to consider the behavior of the absolute value at points where the expression inside is zero. Set \[9-x^2 = 0 \] Solve the equation \[x^2 = 9\]\[x = 3\] or \(x = -3\), both of which are outside our closed interval. Thus, the absolute value doesn't change sign within \([-5, 1]\), and there are no critical points in this interval from the derivative.

Step 2: Evaluate Function at Interval Endpoints

The absolute maximum and minimum could occur at either the critical points within the interval or the endpoints. Evaluate the function at the endpoints:\[f(-5) = 1 + |9 - (-5)^2| = 1 + |9 - 25| = 1 + 16 = 17\]\[f(1) = 1 + |9 - 1^2| = 1 + |9 - 1| = 1 + 8 = 9\]

Step 3: Checking Sign Changes within Interval

Since \(x = 3\) and \(x = -3\) are outside our interval \([-5, 1]\), no further evaluations of \(x\) within \([-5, 1]\) are needed, as \(f(x)\) is continuous and differentiable otherwise in the interval. The absolute value does not cause changes in \(f(x)\)'s behavior in terms of local maxima or minima, as it doesn't toggle within the selected interval.

Step 4: Conclusion on Maximum and Minimum Values

The maximum value calculated is at \(f(-5) = 17\) and the minimum value is at \(f(1) = 9\). Thus, the absolute maximum is 17 at \(x = -5\), and the absolute minimum is 9 at \(x = 1\).

Key Concepts

Absolute Maximum

The term \("absolute maximum"\) refers to the highest value of a function within a specific range or interval. To find this value, we evaluate the function at critical points and endpoints of the interval. In our function, \(f(x)=1+|9-x^{2}|\), the absolute maximum occurs at \(x=-5\) within the closed interval \([-5,1]\). Here, the value of the function is calculated as:

• \(f(-5) = 1 + |9 - (-5)^2| = 1 + 16 = 17\)

Hence, \(x=-5\) is the point where \(f(x)\) reaches its absolute maximum of \(17\) in the interval.

Absolute Minimum

The \("absolute minimum"\) of a function in an interval is the lowest value it takes within that interval. For the function \(f(x)=1+|9-x^2|\), the absolute minimum is found by checking values at the endpoints of the interval, because there are no critical points within our closed interval that yield minimums when the derivative is considered. Inside the given interval \([-5, 1]\), the endpoint \(x=1\) results in:

• \(f(1) = 1 + |9-1^2| = 1 + 8 = 9\)

This makes 9 the absolute minimum value at \(x=1\).

Critical Points

"Critical points" are where the derivative of a function is either zero or undefined, potentially indicating local maxima, minima, or saddle points. In our context:- For the function \(f(x)=1+|9-x^2|\), its critical points would arise by solving \(9-x^2 = 0\), giving \(x=3\) and \(x=-3\).- However, those critical points fall outside the interval \([-5, 1]\), meaning they are irrelevant when seeking absolute extrema within the specified boundaries.

Closed Interval

A "closed interval," such as \([-5, 1]\), includes its endpoints. This is important as it ensures the function's examinations are complete, particularly for finding absolute extrema. By evaluating the function at \(x = -5\) and \(x = 1\), we guarantee capturing the true maximum and minimum, as values within a closed interval cannot exceed evaluations performed at endpoints, given continuity and differentiability.

Derivative

The "derivative" of a function provides valuable insights into its behavior, showing how the function's rate of change varies at different points. For \(f(x) = 1 + |9-x^2|\), we would typically differentiate to find where potential extrema occur. However, the derivative is complicated by the absolute value part. Usually, a derivative close to zero might suggest a critical point. Nonetheless, since \(x = 3\) and \(x = -3\) don't lie in our interval \([-5, 1]\), the derivative does not indicate any critical points within. As a result, we rely on endpoints to find extrema.