Problem 15 - Calculus Early Transcendentals

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: 1 at x = -3.

Step 1: Identify Critical Points

The function is a piecewise function due to the absolute value. First, identify where the function changes. Setting the expression inside the absolute value to zero, we solve: \(9-x^2 = 0\). This gives critical points at \(x = 3\) and \(x = -3\). Only \(x = -3\) is in the interval \([-5, 1]\).

Step 2: Evaluate Function at Critical Point

Substitute the critical point \(x = -3\) into the function. Compute \(f(-3) = 1 + |9 - (-3)^2| = 1 + |9 - 9| = 1\).

Step 3: Evaluate Function at Endpoints

Substitute the interval endpoints into the function: - For \(x = -5\), compute \(f(-5) = 1 + |9 - (-5)^2| = 1 + |9 - 25| = 1 + 16 = 17\).- For \(x = 1\), compute \(f(1) = 1 + |9 - 1^2| = 1 + |9 - 1| = 1 + 8 = 9\).

Step 4: Determine Absolute Maximum and Minimum

Compare the values obtained: \(f(-3) = 1\), \(f(-5) = 17\), and \(f(1) = 9\). The absolute maximum value is \(17\) at \(x = -5\), and the absolute minimum value is \(1\) at \(x = -3\).

Key Concepts

Critical Points

Critical points are where the slope of the function is zero or undefined. These points are important because they help us find where the function could be at a maximum or minimum.

When dealing with functions that have an absolute value, as in the given function, it is vital to know where the inside of the absolute value expression is equal to zero. This change point is a guide to finding critical points. Here, the critical points occur by solving the equation where the expression inside the absolute value becomes zero.

For the function given, we solve the equation \(9 - x^2 = 0\). By solving this, we find that critical points appear at \(x = \pm 3\). Since our interval of interest is \([-5, 1]\), we only consider \(x = -3\) as our valid critical point within the interval.

Examining critical points ensures no maximum or minimum values are missed within the function's range on the interval.

Piecewise Function

A piecewise function is one that has different expressions based on the value of the input. The presence of an absolute value in a function often results in a piecewise function. This is because an absolute value creates two possible cases: either the expression inside the absolute value is positive as it is, or it is negative and becomes positive when multiplied by \(-1\).

In our example, \(f(x) = 1 + |9 - x^2|\) becomes \(1 + (9 - x^2)\) when \(9 - x^2\) is positive and \(1 + (- (9 - x^2))\) when it is negative. These two scenarios make the function piecewise:

• For \(x^2 \leq 9\), \(f(x) = 1 + (9 - x^2)\)
• For \(x^2 > 9\), \(f(x) = 1 - (9 - x^2)\)

Understanding that the absolute value can divide the function into different pieces is crucial. It sets the stage for correctly evaluating outputs at critical points, or where the input changes the nature of the equation.

Evaluation of Endpoints

Evaluating the endpoints of the interval is a decisive step when determining the absolute maximum and minimum values. Endpoints mark the boundaries of what we're interested in, meaning the highest or lowest values of a function often occur with one of them.

The function \(f(x) = 1 + |9 - x^2|\) needs to be evaluated at both \(x = -5\) and \(x = 1\), which are the endpoints of our closed interval \([-5, 1]\). By substituting these endpoints into the function, we get:

• For \(x = -5\), \(f(-5) = 1 + |9 - 25| = 17\)
• For \(x = 1\), \(f(1) = 1 + |9 - 1| = 9\)

By comparing these values with those from any critical points inside the interval, we can confidently determine which are the absolute maximum and minimum.