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. $$f(x)=\left(x^{2}-1\right) \sin ^{-1} x \text { on }[-1,1]$$

Short answer

Question: Verify that the zeros of the derivative of the function \(f(x) = (x^2 - 1)\sin^{-1}(x)\) correspond to points where \(f(x)\) has a horizontal tangent line. Answer: The zeros of the derivative \(f^\prime(x) = (x^2 - 1) \frac{1}{\sqrt{1-x^2}} + 2x\sin^{-1}(x)\) correspond to the points \(x = \pm 1\) and when \(2x\sin^{-1}(x) = (x^2 - 1) \frac{1}{\sqrt{1-x^2}}\). By observing the graphs of \(f(x)\) and \(f^\prime(x)\), we can verify that these zeros indeed correspond to the points where \(f(x)\) has a horizontal tangent line.

Step-by-step solution

Part a: Graphing Function f(x)

Graph the function \(f(x) = (x^2 - 1)\sin^{-1}(x)\) using a graphing utility. You should observe the behavior of the graph on the given interval \([-1,1].\)

Part b: Computing and Graphing the Derivative f'(x)

To compute the derivative \(f^\prime(x)\), we will use the product rule \((uv)^\prime = u^\prime v + uv^\prime\). Let \(u = x^2-1\) and \(v = \sin^{-1}(x)\). First, compute the derivative of \(u\): $$u^\prime = \frac{d}{dx}(x^2-1) = 2x$$ Next, compute the derivative of \(v\). Recall that the derivative of \(sin^{-1}(x)\) is \(\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}\): $$v^\prime = \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}$$ Now, apply the product rule: $$f^\prime(x) = (x^2 - 1) \frac{1}{\sqrt{1-x^2}} + 2x\sin^{-1}(x)$$ To graph the derivative \(f^\prime(x)\), use a graphing utility, and plot the function in the interval \([-1,1].\)

Part c: Verifying Horizontal Tangent Points

To verify that the zeros of \(f^{\prime}(x)\) correspond to points where \(f(x)\) has a horizontal tangent line, first identify where \(f^\prime(x) = 0\): $$(x^2 - 1) \frac{1}{\sqrt{1-x^2}} + 2x\sin^{-1}(x) = 0$$ Analyze the equation, the zeros of the derivative are only possible when either: 1. \(x^2 - 1 = 0 \Rightarrow x = \pm 1\) 2. \(2x\sin^{-1}(x) = (x^2 - 1) \frac{1}{\sqrt{1-x^2}}\) By observing the graph of \(f(x)\) and \(f^\prime(x)\), you will find that the zero of \(f^\prime(x)\) indeed corresponds to the points where \(f(x)\) has a horizontal tangent line. Therefore, our verification is complete.

Key concepts

Product Rule

The **Product Rule** is an essential tool in calculus for finding the derivative of a product of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), their product \( u(x)v(x) \) has a derivative calculated as follows:

• First, take the derivative of the first function \( u(x) \), denoted as \( u'(x) \).
• Multiply \( u'(x) \) by the second function \( v(x) \).
• Then, take the derivative of the second function \( v(x) \), denoted as \( v'(x) \).
• Multiply \( v'(x) \) by the first function \( u(x) \).
• Add these two products together to get the derivative of the entire product: \((uv)' = u'v + uv'\).

In our exercise, the function provided is \( f(x) = (x^2 - 1)\sin^{-1}(x) \). By applying the product rule, you separate this function into \( u(x) = x^2 - 1 \) and \( v(x) = \sin^{-1}(x) \). The calculated derivative then assists in finding the slopes of the function's tangent lines.

Inverse Trigonometric Functions

**Inverse Trigonometric Functions** are functions that are capable of determining the angle given a ratio, essentially reversing what the standard trigonometric functions do. For example, the inverse sine function, denoted as \( \sin^{-1}(x) \), gives you an angle \( \theta \) for which the sine is \( x \).The derivative of \( \sin^{-1}(x) \) is important for calculus applications, such as finding the slope of a tangent line. It is computed as:\[\frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1-x^2}}\]This fractional derivative describes how \( \sin^{-1}(x) \) changes with small changes in \( x \). Note the restriction: this function is only defined for \( x \) in the range of \([-1, 1]\), ensuring that the value under the square root is non-negative. Such restrictions are crucial when working with functions to avoid undefined expressions.

Horizontal Tangent Line

A **Horizontal Tangent Line** occurs at a point on a graph where the slope of the tangent is zero. This means the derivative of the function at this point is zero. A horizontal tangent line suggests a peak, valley, or a flat spot on the curve of the function.To verify where these occur on a function \( f(x) \), you first need to compute its derivative \( f'(x) \) and then find the values of \( x \) for which \( f'(x) = 0 \). This was done in the given exercise:The derivative \( f'(x) \) was calculated as:\[ f'(x) = (x^2 - 1) \frac{1}{\sqrt{1-x^2}} + 2x\sin^{-1}(x) \]Setting \( f'(x) \) to zero allows us to explore:

• Where \( x^2 - 1 = 0 \), simplifying to \( x = \pm 1 \).
• Or where the balance between the terms equalizes, indicating potential unique solutions.

This approach reveals where the original function \( f(x) \) has zero slope and thus forms a horizontal tangent, helping us understand the behavior of the function visually and analytically.