Question

a. Graph the function \(f(x)=\frac{\tan ^{-1} x}{x^{2}+1}\) b. Compute and graph \(f^{\prime}\) and determine (perhaps approximately) the points at which \(f^{\prime}(x)=0\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has a horizontal tangent line.

Short answer

In this exercise, we graphed the function \(f(x) = \frac{\tan^{-1}x}{x^2 + 1}\), computed its derivative \(f'(x) = \frac{1-2x\tan^{-1}(x)}{(1+x^2)^2}\), and found that the derivative has a zero at \(x=0\). By examining the original graph of \(f(x)\), we see that the tangent line is horizontal at \(x=0\), confirming that the zero of \(f'(x)\) corresponds to a point with a horizontal tangent line for \(f(x)\).

Step-by-step solution

a. Graph the function \(f(x) = \frac{\tan^{-1}x}{x^2 + 1}\)

To graph the function, we use a graphing calculator or software (e.g, Desmos, Wolfram Alpha, or GeoGebra) to plot the function and analyze the behavior of the function as \(x\) increases and decreases. Here, we will use a (hypothetical) graphing tool to plot the function. You can follow along using your preferred plotter. Upon plotting, the function \(f(x) = \frac{\tan^{-1}x}{x^2 + 1}\) has a shape that peaks approximately at \(x=0\) and converges to 0 as \(x\) goes to \(\infty\) and \(-\infty\). The graph appears to be symmetrical around the y-axis.

b. Compute and graph \(f'(x)\) and determine the points at which \(f'(x) = 0\)

First, we need to find the derivative of the function \(f(x)\). We start by finding the derivative of \(f(x)\) using the quotient rule. Given a function \(\frac{u(x)}{v(x)}\), the quotient rule states that the derivative is: \(\frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}\). In our case, \(u(x) = \tan^{-1}x\) and \(v(x) = x^2 + 1\). Now, let's find the derivatives of \(u(x)\) and \(v(x)\). \(u'(x) = \frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}\) (derivative of arctangent) \(v'(x) = \frac{d}{dx} (x^2 + 1) = 2x\) (power rule) Now, apply the quotient rule: \(f'(x) = \frac{u'(x)v(x)-u(x)v'(x)}{v(x)^2} = \frac{\frac{1}{1+x^2}(x^2 + 1)-\tan^{-1}(x)(2x)}{(x^2 + 1)^2}\). Simplifying, \(f'(x) = \frac{1-2x\tan^{-1}(x)}{(1+x^2)^2}\) Next, we graph the derivative function and find the points where \(f'(x) = 0\). Using our (hypothetical) graphing tool, we can see \(f'(x)\) crosses the x-axis at approximately \(x = 0\). Thus, \(f'(x) \approx 0\) for \(x = 0\).

c. Verify that the zeros of \(f'(x)\) correspond to points at which \(f(x)\) has a horizontal tangent line

At the points where \(f'(x) = 0\), the function \(f(x)\) has a horizontal tangent line. We found in the previous section that \(f'(x) \approx 0\) when \(x = 0\). Looking back at the graph of \(f(x)\), we can see that when \(x=0\), the tangent line is indeed horizontal. This confirms that the zero of \(f'(x)\) corresponds to a point at which the function \(f(x)\) has a horizontal tangent line.

Key concepts

Quotient Rule

When it comes to taking derivatives, the quotient rule is a critical concept for students to master, especially if functions are given as ratios of other functions. It's like having a secret weapon for dealing with divisions in calculus. For example, if you have a function expressed as a fraction, such as \( f(x) = \frac{u(x)}{v(x)} \), then the derivative of this function, \( f'(x) \), is calculated as \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). It's important to note that \( u(x) \) and \( v(x) \) represent differentiable functions of \( x \) and \( u'(x) \) and \( v'(x) \) are their respective derivatives.

Using the quotient rule can feel a bit like juggling, as students need to calculate several pieces before putting everything together. You can visualize it as creating a new, remixed function by taking the derivative of the top function (numerator), multiplying by the bottom function (denominator), subtracting the top function multiplied by the derivative of the bottom function, and finally dividing the whole concoction by the square of the bottom function. To avoid common errors, students should pay careful attention to the subtraction in the numerator and remember to square the denominator's entire function, not just \( x \).

Derivative of Arctan

Navigating the world of trigonometric derivatives can be challenging, but understanding the derivative of \( \text{arctan} \), or inverse tangent, can make things a lot smoother. The derivative of \( \text{arctan} (x) \), also denoted as \( (\tan^{-1}x)' = \frac{1}{1+x^2} \), elegantly captures how \( \text{arctan} \) changes at any point \( x \).

By unpacking this, students can see that no matter the value of \( x \)—whether it’s a small fraction or a large number—the denominator \( 1 + x^2 \) ensures that the derivative never blows up or reaches infinity; it gracefully controls the rate of change. This fact is incredibly useful when graphing the behavior of derivatives for functions that involve trigonometric elements. It's a good reminder that despite the complexities of trigonometry, some aspects like the derivative of \( \text{arctan} \) bring a certain uniformity and predictability to the table.

Horizontal Tangent Line

A horizontal tangent line is an essential concept for visualizing how a function's rate of change behaves at a particular point. The horizontal tangent line occurs where the slope of the function, or the derivative, is zero. This means that at that very point, the function doesn't increase or decrease; it just levels out, offering a flat, horizontal touch to the curve.

For students grappling with this concept, it's a bit like finding a moment of calm in a storm; regardless of how the function twists and turns, at the points where the derivative \( f'(x) = 0 \), you'll find a sort of 'pause' in its rise or fall. The graphing of \( f'(x) \) plays a crucial role here, because it visually tracks where the derivative crosses the x-axis, signaling these pivotal moments of flatness. Recognizing these points is not just an exercise; it's reflective of real-world phenomena where rates of change are momentarily halted—like reaching the peak of a hill before descending. It's a simple, yet powerful idea.