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

Short answer

Question: Verify that the zeros of the derivative of the function f(x) = (x - 1)sin^(-1)(x) correspond to the points at which f(x) has horizontal tangent lines. Answer: To verify that the zeros of the derivative of the function f(x) = (x - 1)sin^(-1)(x) correspond to the points at which f(x) has horizontal tangent lines, follow these steps: 1. Graph f(x) on the interval [-1, 1] and observe the points with horizontal tangent lines. 2. Find the derivative f'(x) = sin^(-1)(x) + (x-1) (1/√(1-x²)). 3. Graph f'(x) and observe its zeros. 4. Compare the zeros of f'(x) with the points of f(x) with horizontal tangent lines. If they match, the verification is successful.

Step-by-step solution

Graphing the function f(x)

Use a graphing software or calculator, such as Desmos or Wolfram Alpha, to graph the function f(x) = (x - 1)sin^(-1)(x) on the interval [-1, 1]. Observe the shape of the graph and determine the points at which f(x) has horizontal tangent lines.

Find the derivative of f(x)

Next, we need to find the derivative f'(x) of the function. To do this, we must apply the product rule since f(x) is a product of two functions: (x-1) and sin^(-1)(x). The product rule states that the derivative of a product of two functions is: \((u(x)v(x))' = u'(x)v(x) + u(x)v'(x)\) Here, we have: u(x) = x-1 v(x) = sin^(-1)(x) So, find the derivatives of u(x) and v(x): u'(x) = 1 The derivative of the inverse sine function v(x) or arcsin(x) is \(\dfrac{1}{\sqrt{1-x^2}}\). v'(x) = \(\dfrac{1}{\sqrt{1-x^2}}\) Now, apply the product rule and calculate the derivative of f(x): f'(x) = u'(x)v(x) + u(x)v'(x) f'(x) = (1)sin^(-1)(x) + (x-1)\(\dfrac{1}{\sqrt{1-x^2}}\) f'(x) = sin^(-1)(x) + (x-1)\(\dfrac{1}{\sqrt{1-x^2}}\)

Graph the derivative f'(x)

Now, use your graphing tool to graph the first derivative, f'(x) = sin^(-1)(x) + (x-1)\(\dfrac{1}{\sqrt{1-x^2}}\). Observe the zeros of f'(x) and their corresponding x-coordinates.

Verify f'(x) zeros and horizontal tangent lines

Finally, compare the zeros of f'(x) with the points on the graph of f(x) where there are horizontal tangent lines. To do this, look at the graph of f(x) and find the x-coordinates where the tangent lines are horizontal. Make sure these x-coordinates match the zeros of f'(x). If the zeros of f'(x) match the x-coordinates of the points where f(x) has horizontal tangent lines, then we have successfully verified that the zeros of f'(x) correspond to the points at which f(x) has horizontal tangent lines.

Key concepts

Product Rule

Understanding the product rule is fundamental when differentiating functions that are products of two or more functions. When you have a function that consists of two functions multiplied together, like the function given in the exercise (x - 1)sin^-1(x), you cannot simply differentiate each part independently. Instead, you apply the product rule to find the derivative correctly.

The product rule states that for two differentiable functions, u(x) and v(x), the derivative of their product is given by the formula: (u(x)v(x))' = u'(x)v(x) + u(x)v'(x).

To apply this, you should first identify u(x) and v(x), then find their individual derivatives, u'(x) and v'(x). Afterward, you combine them according to the formula above. The product rule is critical for ensuring that your calculus work yields accurate results when dealing with the multiplication of functions.

Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arc-functions, are the inverses of the trigonometric functions and are used to find angle measures when given two sides of a right triangle. In the context of calculus, these functions are often differentiated to analyze the rates of change related to angles.

Some of the most common inverse trigonometric functions are arcsin(x), arccos(x), and arctan(x). The derivative of the arcsin(x) function is particularly relevant to this exercise. It is expressed as v'(x) = 1/ sqrt(1 -

This derivative tells us how fast the angle is changing with respect to changes in the ratio x. It's crucial to be familiar with the derivatives of the inverse trigonometric functions since they appear frequently in calculus problems, as seen in the calculus step involving finding the derivative of a function that includes sin^-1(x). Understanding how to differentiate these functions allows you to solve more complex problems involving angles and their rates of change.

Horizontal Tangent Lines

Horizontal tangent lines are lines that touch a curve at a given point and have a slope of zero. These lines are important in the study of functions, as they can indicate critical points that represent the local maxima and minima or points of inflection on a curve.

To identify the points at which a function f(x) has horizontal tangent lines, you look for where its derivative f'(x) equals zero. These are the points where the instantaneous rate of change of the function is zero, meaning that the function has either a 'flat' peak or trough, or is transitioning from increasing to decreasing or vice versa, without any sharp turn.

In the context of the exercise, by graphing f'(x) and locating its zeros, you can pinpoint exactly where these horizontal tangent lines occur on the original function f(x). Verifying the relationship between the zeros of the derivative and the horizontal tangent lines on the graph of f(x) is a crucial step in understanding the various changes in the function's behavior and how it reflects the function's geometry.