Question

a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=e^{-x} \tan ^{-1} x \text { on }[0, \infty)$$

Short answer

#short_answer# To solve this exercise, we first compute the derivative of the given function f(x) using the product rule since it is a product of two functions, such as e^(-x) and arctan(x). After calculating the derivative, f'(x), we can use a graphing utility to graph both f(x) and f'(x). Next, analyze the zeros of the derivative by comparing the graph of f'(x) or solving f'(x) = 0 algebraically. Finally, verify that these zeros correspond to the points on the graph of f(x) with a horizontal tangent line. If these conditions hold, we have successfully completed the verification process.

Step-by-step solution

Compute the derivative of \(f(x)\)

To find the derivative of \(f(x)\), we need to use the product rule since the given function is a product of two functions (\(e^{-x}\) and \(\tan^{-1}(x)\)). The product rule can be written as: $$(u \cdot v)' = u' \cdot v + u \cdot v'$$ where \(u(x) = e^{-x}\) and \(v(x) = \tan^{-1}(x)\). First, compute the derivatives of \(u(x)\) and \(v(x)\): $$u'(x) = \frac{d}{dx} (e^{-x}) = -e^{-x}$$ $$v'(x) = \frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1+x^2}$$ Now apply the product rule: $$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = -e^{-x} \tan^{-1}(x) + e^{-x} \frac{1}{1+x^2}$$

Graph \(f'(x)\)

You can use a graphing utility to graph the computed derivative function \(f'(x)\). While we can't provide a utility directly here, the steps to input the function are relatively easy. Simply enter the computed \(f'(x)\) equation with the correct syntax for your specific graphing utility.

Verify that zeros of \(f'(x)\) correspond to horizontal tangent lines

To verify that the zeros of the derivative, \(f'(x)\), represent points of horizontal tangent lines on \(f(x)\), recall that a horizontal tangent line occurs when the derivative of a function is equal to zero. This means that, for any point \((x, f(x))\), if \(f'(x) = 0\), then there is a horizontal tangent line at that point. By analyzing the graph of \(f'(x)\) or solving \(f'(x) = 0\) algebraically, we can find the zeros of the derivative. Comparing these zeros with the points on the graph of \(f(x)\) will help us verify that a horizontal tangent line exists at that particular point. If we find that these zeros correspond to those points on the graph of \(f(x)\) with a horizontal tangent line, then we have successfully completed part (c) of this exercise.

Key concepts

Derivative

The derivative is a fundamental tool in calculus that measures how a function changes as its input changes. It gives us the rate of change or the slope of the function at any given point. To compute the derivative of a function like \( f(x) = e^{-x} \tan^{-1}(x) \), we apply certain rules or formulas like the product rule.

The product rule is especially useful when you have a function that is the product of two other functions. In this scenario, those functions are \( e^{-x} \) and \( \tan^{-1}(x) \). Understanding how to apply the product rule is crucial for finding derivatives correctly.

Product Rule

The product rule is a method for finding the derivative of the product of two functions. The rule states that if you have two functions \( u \) and \( v \), their product \( u \cdot v \) has a derivative given by:

• \( (u \cdot v)' = u' \cdot v + u \cdot v' \)

Here, \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

In our problem, \( u(x) = e^{-x} \) and \( v(x) = \tan^{-1}(x) \). We calculate their derivatives:

• \( u'(x) = -e^{-x} \)
• \( v'(x) = \frac{1}{1+x^2} \)

Applying the product rule carefully allows us to find \( f'(x) \), ensuring we handle both parts of the product correctly.

Horizontal Tangent Line

A horizontal tangent line on the graph of a function is where the slope of the function is zero. This usually indicates a local maximum, minimum, or inflection point. For any function \( f(x) \), a horizontal tangent occurs at points where \( f'(x) = 0 \).

To find these points for \( f(x) = e^{-x} \tan^{-1}(x) \), we need to determine where its derivative \( f'(x) \) equals zero. Solving \( -e^{-x} \tan^{-1}(x) + e^{-x} \frac{1}{1+x^2} = 0 \) gives us the x-values for where the tangent to the curve is horizontal. At these points, \( f(x) \) does not increase or decrease, which is a key concept in critical point analysis.

Graphing Utility

A graphing utility is an essential tool that allows us to visualize functions and their derivatives, helping us to better understand their behavior. With modern technology, graphing utilities come in the form of software, calculators, and online tools.

To examine the function \( f(x) = e^{-x} \tan^{-1}(x) \) and its derivative \( f'(x) \), you input the equations into the graphing utility. This helps identify the points where \( f'(x) \) equals zero. Such visualization confirms the analytical work done by showing the corresponding horizontal tangent lines on the graph of \( f(x) \).

• Graph the original function for rough analysis.
• Graph the derivative to identify points with zero slopes.
• Check for corresponding horizontals on the original graph.

This process ensures the coordinate pairs match, providing a clear view of where the function changes direction or has critical points.