Problem 197 - Calculus Volume 1

Question

Find all \(x\) values on the graph of \(f(x)=-3 \sin x \cos x\) where the tangent line is horizontal.

Step-by-Step Solution

Answer

The \(x\)-values are \(x = \frac{(2n+1)\pi}{4}\), where \(n\) is an integer.

Step 1: Understand the Problem

We need to find the values of \(x\) where the tangent line to the graph of the function \(f(x) = -3 \sin x \cos x\) is horizontal. A horizontal tangent line occurs where the derivative is zero.

Step 2: Find the Derivative

First, use the product rule to find the derivative of \(f(x) = -3 \sin x \cos x\). The product rule states that \((uv)' = u'v + uv'\). Let \(u = -3 \sin x\) and \(v = \cos x\), then \(u' = -3 \cos x\) and \(v' = -\sin x\). Thus, the derivative \(f'(x) = uv' + u'v\) becomes:\[f'(x) = (-3 \sin x)(-\sin x) + (-3 \cos x)(\cos x) = 3 \sin^2 x - 3 \cos^2 x\]

Step 3: Simplify the Derivative

Recognize that the expression \(3 \sin^2 x - 3 \cos^2 x\) can be rewritten using the double angle formula for cosine, \(\cos(2x) = \cos^2 x - \sin^2 x\). This allows simplifying the expression to:\[f'(x) = 3(\sin^2 x - \cos^2 x) = -3(\cos(2x))\]

Step 4: Set the Derivative to Zero

To find where the tangent line is horizontal, set the derivative equal to zero:\[-3 \cos(2x) = 0\]

Step 5: Solve for x

First, recognize that \(-3\) does not affect the equality, so we have:\[\cos(2x) = 0\]The cosine function equals zero at odd multiples of \(\frac{\pi}{2}\), so \(2x = (2n+1)\frac{\pi}{2}\), where \(n\) is any integer. Solving for \(x\) gives:\[x = \frac{(2n+1)\pi}{4}\]

Step 6: Conclude the Solution

The solution indicates that the \(x\)-values for which the tangent line of the function \(f(x) = -3 \sin x \cos x\) is horizontal are given by \(x = \frac{(2n+1)\pi}{4}\) for integers \(n\). These represent all odd multiples of \(\frac{\pi}{4}\), i.e., \(x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \ldots\)

Key Concepts

Product Rule in Calculus

In calculus, the product rule is a fundamental technique for finding the derivative of a product of two functions. Suppose you have a function that is the product of two simpler functions: say, \( u(x) \) and \( v(x) \). The product rule states that the derivative of this product, denoted as \((uv)' \), is \( u'v + uv' \).
This means that to find the derivative, you'll take the derivative of the first function and multiply it by the second, plus the first function itself multiplied by the derivative of the second.In our problem, the function \( f(x) = -3 \sin x \cos x \) is a product of two functions: \( u(x) = -3 \sin x \) and \( v(x) = \cos x \). Thus the product rule helps us find the derivative \( f'(x) \), which is crucial for determining when the tangent line is horizontal. Calculating these derivatives, we get \( u' = -3 \cos x \) and \( v' = -\sin x \).
Applying the product rule, the derivative becomes \( f'(x) = (-3 \sin x)(-\sin x) + (-3 \cos x)(\cos x) = 3 \sin^2 x - 3 \cos^2 x \).

Trigonometric Functions

Trigonometric functions such as sine and cosine are periodic and oscillate between -1 and 1. They are vital in various calculus problems, often requiring specific identities for simplification. In this exercise, both \( \sin x \) and \( \cos x \) are involved, and their squared terms appear in the expression \( 3 \sin^2 x - 3 \cos^2 x \).
One key identity worth knowing is the double angle formula for cosine: \( \cos(2x) = \cos^2 x - \sin^2 x \).
This identity is handy for simplifying expressions involving trigonometric squares.By recognizing this identity, you simplify the derivative to \( f'(x) = -3(\cos(2x)) \). This kind of manipulation makes solving the problem much easier because it transitions the question into a well-known form that is simpler to solve or analyze.

Derivative and Critical Points

When a function's derivative equals zero, it reveals important points called critical points. In practical terms, finding these points tells you where potential local maxima, minima, or inflection points occur.
For this problem, a critical point is where the tangent line becomes horizontal. This implies a slope of zero, thus \( f'(x) = 0 \). Solving \( -3 \cos(2x) = 0 \), simplifies to \( \cos(2x) = 0 \).The cosine function is zero at odd multiples of \( \frac{\pi}{2} \), giving \( 2x = (2n+1)\frac{\pi}{2} \) for integers \( n \). Solving for \( x \) provides \( x = \frac{(2n+1)\pi}{4} \).

• This equation gives all the \( x \) values where the horizontal tangent occurs for the function \( f(x) = -3 \sin x \cos x \).
• These points \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \ldots \) are all odd multiples of \( \frac{\pi}{4} \).

Understanding how derivatives and critical points relate helps in analyzing the behavior of a function and predicting its graph's features.