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. $$f(x)=\left(\sec ^{-1} x\right) / x \text { on }[1, \infty)$$

Short answer

(b) Compute the derivative of the function, \(f'(x)\), and graph it. What do you notice about the zeros of \(f'(x)\)? (c) Verify that the zeros of \(f'(x)\) correspond to points where \(f(x)\) has horizontal tangent lines.

Step-by-step solution

Graphing the function \(f(x)\)

To graph the function \(f(x) = \frac{\sec^{-1}x}{x}\), we can use any graphing utility software or online tool, like Desmos or GeoGebra. The graph should show the behavior of the function on the domain \([1,\infty)\).

Compute the derivative \(f'(x)\) and graph it

To compute the derivative \(f'(x)\), we first need to find the derivative of \(\sec^{-1}(x)\). Recall that the derivative of an inverse function is given by: $$\frac{d}{dx}\left(\sec^{-1}x\right) = \frac{1}{|x|\sqrt{x^2-1}}$$ Now, we'll apply the quotient rule to find the derivative of the function \(f(x)\): $$f'(x) = \frac{\frac{d(\sec^{-1}x)}{dx} \cdot x - \sec^{-1}x\cdot \frac{d(x)}{dx}}{x^2}$$ Substituting the derivative of \(\sec^{-1}x\): $$f'(x) = \frac{\frac{1}{|x|\sqrt{x^2-1}}x- \sec^{-1}x}{x^2}$$ Now, graph the function \(f'(x)\) using a graphing utility software or online tool, like Desmos or GeoGebra.

Verify horizontal tangents

Finally, we need to verify that the zeros of \(f'(x)\) correspond to points where \(f(x)\) has a horizontal tangent. A horizontal tangent occurs when the derivative of the function (\(f'(x)\)) is equal to zero. We already have the expression for \(f'(x)\): $$f'(x) = \frac{\frac{1}{|x|\sqrt{x^2-1}}x- \sec^{-1}x}{x^2}$$ To find the zeros of \(f'(x)\), we set \(f'(x)\) to zero and solve for x: $$\frac{\frac{1}{|x|\sqrt{x^2-1}}x- \sec^{-1}x}{x^2} = 0$$ This equation is quite hard to solve algebraically, but we can find the zeros by analyzing the graph of \(f'(x)\). Locate the zeros of \(f'(x)\) on its graph and observe the corresponding points on the graph of \(f(x)\). If the tangent lines at these points on the graph of \(f(x)\) are horizontal, then it verifies that the zeros of \(f'(x)\) correspond to points at which \(f(x)\) has a horizontal tangent line.

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse trigonometric functions, such as sine, cosine, and tangent. The function \(\sec^{-1}(x)\) is an inverse trigonometric function, specifically the inverse of the secant function. In simple terms, it changes the result of a secant function back to the angle that originally produced it.

These functions are crucial when solving trigonometric equations where we want to find the angle itself instead of the ratio. In the context of our exercise with \(f(x) = \frac{\sec^{-1}x}{x}\), the function \(\sec^{-1}x\) allows us to work with angles as output, which helps in the differentiation process needed later on.

Understanding inverse trigonometric functions helps us graph and analyze behaviors of trigonometric relationships, making it easier to handle problems involving angles and their measurements.

Derivative Computation

The derivative of a function tells us the rate at which the function's value is changing at any point. To compute the derivative of the given function \(f(x) = \frac{\sec^{-1}x}{x}\), we first need to understand the differentiation of \(\sec^{-1}(x)\).

Using the formula, the derivative of \(\sec^{-1}(x)\) is given by \(\frac{1}{|x|\sqrt{x^2-1}}\). Then, we apply the quotient rule to \(f(x)\). The quotient rule helps us differentiate functions that are fractional.

• The quotient rule states: If you have a function \(g(x)/h(x)\), then its derivative is \[\frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}\].
• Applying it to \(f(x)\), we get the derivative \(f'(x)\) as \[f'(x) = \frac{\frac{1}{|x|\sqrt{x^2-1}} \cdot x - \sec^{-1}x}{x^2}\].

This computation is essential for finding where the function's graph has special features, like horizontal tangents, or to understand its slope at different points.

Horizontal Tangent Lines

Horizontal tangent lines occur where the derivative of a function equals zero. This is because a zero derivative means the slope is zero, corresponding to a horizontal line at that point.

In our exercise, after computing the derivative \(f'(x)\), we set it to zero to find potential points of horizontal tangents:

\[\frac{\frac{1}{|x|\sqrt{x^2-1}}x- \sec^{-1}x}{x^2} = 0\]

While this equation might be difficult to solve algebraically, graphing the derivative function \(f'(x)\) on tools like Desmos helps us see where it intersects the x-axis (where the derivative is zero). These points on the x-axis correspond to horizontal tangents in the original function \(f(x)\).

Finding these points is valuable because they often highlight points of interest, such as potential maximums, minimums, or inflection points on the graph of \(f(x)\). Thus, understanding horizontal tangents provides deeper insight into the behavior of functions.