Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x)=x-\sin x\). Find all points on the graph of \(y=f(x)\) where the tangent line is horizontal. Find all points on the graph of \(y=f(x)\) where the tangent line has slope 2 .

Step-by-Step Solution

Answer

Horizontal tangent: points \((2n\pi, 2n\pi)\). Slope 2: points \(((2n+1)\pi, (2n+1)\pi)\).

Step 1: Derivative of the Function

To find the points where the tangent line is horizontal or has a specific slope, we need to calculate the derivative of the function, which gives the slope of the tangent line at any point on the function. The function given is \( f(x) = x - \sin x \). The derivative of \( f(x) \) is:\[ f'(x) = \frac{d}{dx}(x - \sin x) = \frac{d}{dx}(x) - \frac{d}{dx}(\sin x) = 1 - \cos x \].

Step 2: Horizontal Tangent Line Condition

For the tangent line to be horizontal, the slope must be zero. Therefore, set the derivative equal to zero:\[ f'(x) = 1 - \cos x = 0 \]. Solve for \( x \):\[ \cos x = 1 \].The solutions to \( \cos x = 1 \) are \( x = 2n\pi \), where \( n \) is an integer.

Step 3: Slope of 2 Condition

To find where the tangent line has a slope of 2, set the derivative equal to 2:\[ f'(x) = 1 - \cos x = 2 \].Solve for \( x \):\[ -\cos x = 1 \quad \Rightarrow \quad \cos x = -1 \].The solutions to \( \cos x = -1 \) are \( x = (2n + 1)\pi \), where \( n \) is an integer.

Step 4: Determine Corresponding Points on the Graph

Now we know that for a horizontal tangent, \( x = 2n\pi \). The corresponding \( y \)-values are, using \( f(x) = x - \sin x \),\[ y = 2n\pi - \sin(2n\pi) = 2n\pi \].So the points are \((2n\pi, 2n\pi)\).For a tangent slope of 2, \( x = (2n + 1)\pi \). The corresponding \( y \)-values are,\[ y = (2n + 1)\pi - \sin((2n + 1)\pi) = (2n + 1)\pi \].Since \( \sin((2n + 1)\pi) = 0 \), the points are \(((2n+1)\pi, (2n+1)\pi)\).

Key Concepts

Derivative Calculation

Derivative calculation is a fundamental technique in calculus used to find the slope of the tangent line at any point on a curve. In simple terms, the derivative tells us how a function changes as we modify its input. To obtain the derivative, we apply rules of differentiation to the function. For the function given, \( f(x) = x - \sin x \), we differentiate each term separately:

• The derivative of \( x \) is 1.
• The derivative of \( -\sin x \) is \(-\cos x\) as differentiating \( \sin x \) gives \( \cos x \), and multiplying by the negative sign yields \(-\cos x\).

Hence, the derivative of the function is \( f'(x) = 1 - \cos x \). This expression is crucial as it represents the slope of the tangent line to the curve at any point \( x \). Understanding derivatives allows us to explore various properties like tangents, rates of change, and stability of a function.

Horizontal Tangent

A tangent line is horizontal when its slope is zero. This concept is crucial for identifying flat spots on a curve. To find such points, we set the derivative equal to zero and solve for the variable.In our example, we solve the equation:\[ f'(x) = 1 - \cos x = 0 \]This simplifies to \( \cos x = 1 \). The cosine function equals one at even multiples of \( \pi \) (i.e., \( 0, 2\pi, 4\pi, \ldots \)), or more generally \( x = 2n\pi \) where \( n \) is an integer. Each solution corresponds to a flat area on the curve. To find the specific points, substitute back into the original function:

• The \( y \)-value at \( x = 2n\pi \) is \( f(2n\pi) = 2n\pi - \sin(2n\pi) = 2n\pi \) because \( \sin(2n\pi) = 0 \).

Therefore, the points \( (2n\pi, 2n\pi) \) are where the tangent is horizontal.

Slope of Tangent Line

The slope of a tangent line gives us the steepness and direction of this line at any given point on the curve. For this problem, we're asked to find where the slope is 2.We do this by setting the derivative equal to 2:\[ f'(x) = 1 - \cos x = 2 \]Rearranging, we get \( \cos x = -1 \). The cosine of an angle is \(-1\) at odd multiples of \( \pi \): \( \pi, 3\pi, 5\pi, \ldots \), or \( x = (2n+1)\pi \), where \( n \) is an integer.At these \( x \)-values, the function's curve is sloping upward with a slope of 2. Substituting back into \( f(x) \):

• We find \( y \)-values of \( (2n+1)\pi - \sin((2n+1)\pi) = (2n+1)\pi \) as \( \sin((2n+1)\pi) = 0 \).

Thus, the points \( ((2n+1)\pi, (2n+1)\pi) \) indicate where the slope of the tangent line is precisely 2.

Trigonometric Solutions

Trigonometric solutions involve solving equations where trigonometric functions like sine and cosine appear. In this exercise, we deal with cosine functions to determine particular values of \( x \) for which the derivative's conditions are met.When solving \( \cos x = 1 \) or \( \cos x = -1 \), it's useful to remember:

• \( \cos x = 1 \) at frequencies like \( 0, 2\pi, 4\pi \), given by \( x = 2n\pi \).
• \( \cos x = -1 \) at frequencies like \( \pi, 3\pi, 5\pi \), given by \( x = (2n+1)\pi \).

These precise angles reflect inherent periodic properties of the cosine function, arising from its wave-like behavior. Understanding these solutions helps us identify specific points on a graph where certain conditions, such as a horizontal tangent or specified slope, are fulfilled. This knowledge establishes a foundation for exploring more complex trigonometric relationships in calculus.