Problem 20 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. Determine whether or not \(f\) attains an absolute maximum and absolute minimum value. If it does, determine the absolute maximum and/or minimum value. f(x)=\frac{10 x}{x^{2}+1}

Step-by-Step Solution

Answer

The critical points are \(x= \sqrt{0.5}\) and \(x= -\sqrt{0.5}\). After plugging these values along with the limit values into the function, the results will give the absolute maximum and minimum.

Step 1: Find the derivative of f

Use the quotient rule which states that \((u/v)' = (u'v - uv')/v^{2}\). Here, \(u = 10x\) and \(v = x^{2} + 1\). Thus, \(u' = 10\) and \(v' = 2x\). Substituting into the quotient rule gives \(f'(x) = (10(x^{2} + 1) - 10x(2x))/ (x^{2} + 1)^{2} = (10 - 20x^{2})/(x^{2} + 1)^{2}\).

Step 2: Find the critical points

Set the derivative equal to zero (0) and solve for x. \(0 = (10 - 20x^{2}),\ giving x = \pm \sqrt{10/20} = \pm \sqrt{0.5}\). So the critical points are \(\sqrt{0.5}\) and \(-\sqrt{0.5}\).

Step 3: Identify and classify the critical points

Find the second derivative \(f''(x)\) by differentiating \(f'(x)\) again, using the quotient rule as before. Give that \(f''(x) =-40x(x^2 + 1)^{-2} + 80x^2(x^2 + 1)^{-3}\). Substitute \(x=\sqrt{0.5}\) and \(x= -\sqrt{0.5}\) to determine classifications of critical points. If \(f''(x) > 0\), the critical point is a local minimum; if \(f''(x) < 0\), the critical point is a local maximum.

Step 4: Find absolute maximum and minimum

To determine the absolute maximum and minimum, we plug the critical points and the values at the limit of the domain into the original function \(f(x)\) and see the corresponding outputs. Given that the domain of a quadratic function is the set of all real numbers, we plug in \(\pm \sqrt{0.5}\) and \(\pm \infty\) into the function \(f(x)\), then compare the results. The highest value will be the absolute maximum and the lowest value will be the absolute minimum.

Key Concepts

Quotient Rule Differentiation

Differentiation is a fundamental concept in calculus used to find the rate of change of a function. When dealing with the division of two functions, the quotient rule is a handy tool. Mathematically, the quotient rule is expressed as \( (\frac{u}{v})' = \frac{u'v - uv'}{v^{2}} \) where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives.

In the context of our exercise, we applied the quotient rule to differentiate \( f(x) = \frac{10x}{x^{2}+1} \). By identifying \( u = 10x \) and \( v = x^{2} + 1 \) and then finding \( u' = 10 \) and \( v' = 2x \) we could calculate the derivative \( f'(x) \) which is essential in finding critical points. The quotient rule is particularly useful because it avoids the more complex and error-prone method of first applying the product rule to \( uv^{-1} \) and then simplifying.

Absolute Maximum and Minimum

The concepts of absolute maximum and minimum are critical in understanding the behavior of functions over their entire domain. An absolute maximum is the highest point on the graph of a function, whereas an absolute minimum is the lowest point. These points are significant because they signify the extrema or most extreme values that a function can achieve.

To find these values, as in our exercise, we need to consider both the critical points and the limits as \( x \) approaches the boundaries of the domain. For the function \( f(x) = \frac{10x}{x^{2}+1} \), the domain is all real numbers. After finding the critical points by setting the derivative equal to zero, we also consider the behavior of \( f(x) \) as \( x \) approaches infinity, which can represent potential absolute extrema. By evaluating \( f(x) \) at these points, the highest and lowest outputs are determined as the absolute maximum and minimum.

Second Derivative Test

The second derivative test is a convenient method for classifying critical points as local maxima, minima, or points of inflection. Once the critical points are found, their classification depends on the concavity of the function at those points, which is determined by the second derivative \( f''(x) \).

If \( f''(x) > 0 \) at a critical point, the function is concave up, and the point is a local minimum. Conversely, if \( f''(x) < 0 \) at a critical point, the function is concave down, and the point is a local maximum. For the function \( f(x) \) from our exercise, the second derivative is calculated to assess the critical points found earlier. By plugging in the values of the critical points into the second derivative, we can classify them appropriately, helping us to better understand the function's overall shape and potentially identify its absolute maximum and minimum.