Question

a. Locate the critical points of \(f\) b. Use the First Derivative Test to locate the local maximum and minimum values. c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist). $$f(x)=\frac{x^{2}}{x^{2}-1} \text { on }[-4,4]$$

Short answer

Answer: The absolute maximum value is 4 at \(x = 2\), and the absolute minimum value is approximately 1.882 at \(x = -4\) and \(x = 4\).

Step-by-step solution

Find the first derivative of the function

Differentiate the given function with respect to x. We can use the quotient rule to find the derivative: $$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} $$ Here, \(u = x^2\) and \(v = x^2 - 1\), so: $$ \frac{du}{dx} = 2x \quad \text{ and } \quad \frac{dv}{dx} = 2x $$ Now using the quotient rule, we get: $$ f'(x) = \frac{(x^2 - 1)2x - x^2(2x)}{(x^2-1)^2} = \frac{2x^3-2x^4-2x^3}{(x^2-1)^2} $$ Now, simplify the expression: $$ f'(x) = \frac{-2x^4+4x^3}{(x^2-1)^2} $$

Locate the critical points

To find the critical points, we need to solve the equation \(f'(x) = 0\) and find points where the derivative is undefined. Setting the numerator of the derivative equal to zero, we get: $$ -2x^4+4x^3 = 0 $$ Factor the equation: $$ -2x^3(x-2) = 0 $$ This gives us two critical points, \(x=0,2\). The points where the derivative is undefined occur when the denominator is equal to zero: $$ (x^2-1)^2 = 0 $$ Which implies \(x=1, -1\). Since these points are undefined, they are not critical points.

Apply the First Derivative Test

Using the First Derivative Test, classify the critical points as local maxima, minima, or neither. To do this, pick test points in the intervals: \((-4, 0)\), \((0, 2)\), and \((2, 4)\). For the interval \((-4, 0)\), let's choose \(x = -2\). Plugging into the first derivative \(f'(x)\) we have: $$ f'(-2) = \frac{-2(-2)^4+4(-2)^3}{((-2)^2-1)^2} > 0 $$ Since \(f'(x)\) is positive in this interval, the function is increasing and we have a local minimum at \(x=0\). For the interval \((0, 2)\), let's choose \(x = 1\). f'(x) is undefined at x=1, so we can't apply the First Derivative Test here. For the interval \((2, 4)\), let's choose \(x = 3\). Plugging into the first derivative \(f'(x)\) we have: $$ f'(3) = \frac{-2(3)^4+4(3)^3}{((3)^2-1)^2} < 0 $$ Since \(f'(x)\) is negative in this interval, the function is decreasing and we have a local maximum at \(x=2\).

Identify the absolute maximum and minimum values

To identify the absolute maximum and minimum values on the interval \([-4, 4]\), we need to consider the local extrema (found in Step 3) and the endpoints of the interval. Evaluate the function for the critical points and the endpoints: $$ f(-4) = \frac{(-4)^2}{(-4)^2 - 1} \approx -1.882 \\ f(0) = \frac{0^2}{0^2 - 1} = 0\\ f(2) = \frac{2^2}{2^2 - 1} = 4 \\ f(4) = \frac{(4)^2}{(4)^2 - 1} \approx 1.882 $$ The absolute maximum value occurs at \(x = 2\) with \(f(2) = 4\). The absolute minimum value occurs at \(x = -4\) and \(x = 4\) with \(f(-4) \approx f(4) \approx 1.882\).

Key concepts

Critical Points

Critical points are the values of \(x\) where the first derivative of a function equals zero or is undefined. These points are essential as they indicate where the function could potentially have a local maximum or minimum. To find the critical points, we need to follow a couple of steps:

• Calculate the derivative \(f'(x)\).
• Solve \(f'(x) = 0\) to identify where the slope of the function equals zero.
• Check where \(f'(x)\) is undefined, as these can also signal interesting behavior in the function.

In our example with the function \(f(x) = \frac{x^2}{x^2 - 1}\), we applied these methods to determine critical points at \(x = 0\) and \(x = 2\). Although \(x = 1\) and \(x = -1\) make the derivative undefined, these are not classified as critical points because the function itself isn’t defined there.
This is crucial for understanding potential maxima and minima.

First Derivative Test

The First Derivative Test is a technique used to classify critical points further into local maxima, minima, or saddle points. After locating the critical points, the following steps are used:

• Choose test points in the intervals formed by the critical points.
• Plug these test points into \(f'(x)\) to determine the sign of the derivative.
• If \(f'(x)\) changes from positive to negative at a critical point, it's a local maximum; from negative to positive, it's a local minimum.
• If the sign doesn't change, the critical point is not a local extremum.

For the function \(f(x) = \frac{x^2}{x^2 - 1}\), by testing values in each interval around our critical points \(x = 0\) and \(x = 2\), we found a local minimum at \(x = 0\) and a local maximum at \(x = 2\). This classification helps in understanding the behavior and shape of the function around critical points.

Absolute Maximum and Minimum Values

Absolute maximum and minimum values refer to the greatest and smallest values, respectively, that a function reaches on a given interval. To identify them, we evaluate the function not only at critical points but also at the endpoints of the interval. The process involves:

• Calculating \(f(x)\) at critical points within the interval.
• Calculating \(f(x)\) at the endpoints of the interval.
• Comparing these values to determine the highest and lowest function values.

For the function \(f(x) = \frac{x^2}{x^2 - 1}\) on the interval \([-4, 4]\), we evaluated \(f(x)\) at the critical points \(x = 0\) and \(x = 2\) as well as at the endpoints \(x = -4\) and \(x = 4\). The analysis showed that the absolute maximum value of 4 occurs at \(x = 2\), while the absolute minimum values, approximately \(1.882\), are at the endpoints \(x = -4\) and \(x = 4\). This method provides a comprehensive view of the function’s range on a specified interval.