Question

A family of double-humped functions Consider the functions \(f(x)=\frac{x}{\left(x^{2}+1\right)^{n}},\) where \(n\) is a positive integer. a. Show that these functions are odd for all positive integers \(n .\) b. Show that the critical points of these functions are \(x=\pm \frac{1}{\sqrt{2 n-1}},\) for all positive integers \(n .\) (Start with the special cases \(n=1\) and \(n=2 .\) ) c. Show that as \(n\) increases, the absolute maximum values of these functions decrease. d. Use a graphing utility to verify your conclusions.

Short answer

#Question# Prove that the double-humped functions defined by \(f(x)=\frac{x}{\left(x^{2}+1\right)^{n}}\), where \(n\) is a positive integer, are odd.

Step-by-step solution

Part a: Proving that the functions are odd

To prove that a function is odd, we need to show that \(f(-x)=-f(x)\) for all \(x.\) Let's plug in \(-x\) into the function and see what we get: $$ f(-x)=\frac{-x}{\left((-x)^{2}+1\right)^{n}}=\frac{-x}{\left(x^{2}+1\right)^{n}}=-\frac{x}{\left(x^{2}+1\right)^{n}}=-f(x) $$ Thus, the functions are odd for all positive integers \(n.\)

Part b: Finding the critical points of the function

To find the critical points of a function, we need to find the points where the derivative is equal to zero or is undefined. First, let's find the derivative of the function using the quotient rule: $$ f'(x) = \frac{(x^2 + 1)^n(1) - x(2nx^2(x^2 + 1)^{n-1})}{(x^2 + 1)^{2n}}\\ f'(x) = \frac{(x^2 + 1)^{n-1}(1 - 2nx^2)}{(x^2 + 1)^{2n}} $$ Now, set the numerator equal to zero and solve for \(x\): $$ 1 - 2nx^2 = 0 \Rightarrow x^2 = \frac{1}{2n} \Rightarrow x = \pm\frac{1}{\sqrt{2n}} $$ Thus, the critical points of the function are \(x = \pm\frac{1}{\sqrt{2n}}\), for all positive integers \(n.\)

Part c: Showing that the maximum values decrease as n increases

To find the maximum points, we need to evaluate the function at the critical points we found in part b. We know that as a function is odd, the maximum and minimum values will be at symmetrical points. Thus, we just need to prove that the maximum value at \(x = \frac{1}{\sqrt{2n}}\) decreases as n increases. Let's plug in the critical point into the function: $$ f\left(\frac{1}{\sqrt{2n}}\right) = \frac{\frac{1}{\sqrt{2n}}}{\left(\frac{1}{2n}+1\right)^{n}}= \frac{1}{(\sqrt{2n})\left(1+\frac{1}{2n}\right)^{n}} $$ Now, let's consider the limit as n approaches infinity: $$ \lim_{n \to \infty} \frac{1}{(\sqrt{2n})\left(1+\frac{1}{2n}\right)^{n}}=0 $$ The limit tells us that as n increases, the maximum value of the function approaches zero. Hence, the absolute maximum values of these functions decrease as \(n\) increases.

Part d: Using a graphing utility to verify conclusions

To verify our conclusions in parts a, b, and c, use a graphing utility like Desmos or Grapher. Plot the function \(f(x)=\frac{x}{\left(x^{2}+1\right)^{n}}\) for different values of n and observe the following: 1. For all values of n, the function is odd, meaning it is symmetric about the origin. 2. The critical points (maximum and minimum) are located at \(x = \pm\frac{1}{\sqrt{2n}}\). 3. As n increases, the maximum and minimum values of the function become smaller and approach zero, confirming that the absolute maximum values decrease as n increases.

Key concepts

Understanding Odd Functions

An odd function is defined by its symmetry about the origin, which means that if you have a function like the one given, say,
\( f(x) = \frac{x}{(x^2+1)^n} \),
for it to be considered odd, it should satisfy the condition
\( f(-x) = -f(x) \),
for all values of x. When we substitute \( -x \) for \( x \) in the given function, we end up with the negative of the original function for any value of n, thus confirming its odd nature.

In the context of the double-humped function, this property dictates the function's graphical representation, where one 'hump' will be above the x-axis and the other symmetrically below it. Understanding this helps students visualize the function, aiding their comprehension of the subsequent properties, such as critical points and maximum values.

Identifying Critical Points

Critical points are essential in calculus because they represent where the function's derivative is zero or undefined, often indicating peaks, troughs, or inflection points on the graph.

For the family of functions
\( f(x) = \frac{x}{(x^2+1)^n} \),
we determine the derivative using the quotient rule. The quotient rule, which allows us to differentiate ratios of functions, tells us that the derivative of a fraction
\( \frac{u(x)}{v(x)} \)
is
\( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).
When we apply this to our function and set the numerator equal to zero, we find the critical points at
\( x = \pm\frac{1}{\sqrt{2n}} \).
These points are vital for locating local maxima and minima, which are pivotal in understanding the behavior of the function across its domain.

Investigating Maximum Values of Functions

In the study of functions, maximum values are often of interest because they represent the highest points on a graph. For an odd function like ours, maximum and minimum values will be symmetrical about the origin.

The function's maximum value occurs at the critical point
\( x = \frac{1}{\sqrt{2n}} \),
and as we show in the solution, this value decreases with increasing n. This is because the denominator of the function increases faster than the numerator as n grows, pushing the maximum value closer to zero. Recognizing this trend of decreasing maximum values helps students grasp the impact of the parameter n on the overall shape and height of the graph.

Applying the Quotient Rule in Calculus

The quotient rule in calculus is a technique for differentiating functions that are expressed as the ratio of two differentiable functions.

In our exercise, the function
\( f(x) = \frac{x}{(x^2+1)^n} \)
is such a function. To find its derivative, we identify the numerator as
\( u(x) = x \)
and the denominator as
\( v(x) = (x^2+1)^n \).
With the quotient rule, we can efficiently determine where the function's slope changes—which in turn reveal its critical points. It's important for students to practice the quotient rule because it appears frequently in calculus problems and is instrumental in analyzing the intricate behaviors of complex functions.