What is the slope of the tangent of y = cos -1 (cos x) at x = \(-\frac{\pi}{5}\) ?
-1
Understanding the Function y = cos⁻¹(cos x)
The given function is \( y = \cos^{-1}(\cos x) \). To find the slope of the tangent at a specific point, we first need to understand how this function behaves. The range of the principal value function \( \cos^{-1}(u) \) is \( [0, \pi] \). This means that \( \cos^{-1}(\cos x) \) will always output a value between \( 0 \) and \( \pi \).
The simplification of \( \cos^{-1}(\cos x) \) depends on the interval of \( x \):
- If \( x \in [0, \pi] \), then \( \cos^{-1}(\cos x) = x \).
- If \( x \in [-\pi, 0] \), then \( -x \in [0, \pi] \). We know that \( \cos(-x) = \cos x \). So, \( \cos^{-1}(\cos x) = \cos^{-1}(\cos(-x)) \). Since \( -x \) is in the range \( [0, \pi] \), \( \cos^{-1}(\cos(-x)) = -x \). Thus, for \( x \in [-\pi, 0] \), \( y = -x \).
- For other intervals, the function follows a pattern. For example, if \( x \in [\pi, 2\pi] \), then \( x - 2\pi \in [-\pi, 0] \). \( \cos(x) = \cos(x - 2\pi) \). So, \( \cos^{-1}(\cos x) = \cos^{-1}(\cos(x - 2\pi)) \). Since \( x - 2\pi \in [-\pi, 0] \), \( \cos^{-1}(\cos(x - 2\pi)) = -(x - 2\pi) = 2\pi - x \). Thus, for \( x \in [\pi, 2\pi] \), \( y = 2\pi - x \).
Evaluating the Function at x = -\(\frac{\pi}{5}\)
We are interested in the slope of the tangent at \( x = -\frac{\pi}{5} \). Let's determine which interval this value falls into.
The value \( x = -\frac{\pi}{5} \) is between \( -\pi \) and \( 0 \), specifically \( -\pi < -\frac{\pi}{5} < 0 \).
Since \( x = -\frac{\pi}{5} \) is in the interval \( [-\pi, 0] \), the function \( y = \cos^{-1}(\cos x) \) simplifies to \( y = -x \) for values of \( x \) around \( -\frac{\pi}{5} \).
Finding the Slope of the Tangent
The slope of the tangent to the curve \( y = f(x) \) at a point \( x_0 \) is given by the derivative \( \frac{dy}{dx} \) evaluated at \( x = x_0 \).
For \( x \) values around \( -\frac{\pi}{5} \), the function is \( y = -x \).
Let's find the derivative of \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx}(-x) \)
\( \frac{dy}{dx} = -1 \)
The derivative is a constant value of \( -1 \) for all \( x \) in the interval \( (-\pi, 0) \).
Calculating the Slope at x = -\(\frac{\pi}{5}\)
The slope of the tangent at \( x = -\frac{\pi}{5} \) is the value of the derivative \( \frac{dy}{dx} \) at \( x = -\frac{\pi}{5} \).
Since \( \frac{dy}{dx} = -1 \) for all \( x \) in the relevant interval including \( -\frac{\pi}{5} \), the slope at \( x = -\frac{\pi}{5} \) is \( -1 \).
Thus, the slope of the tangent to \( y = \cos^{-1}(\cos x) \) at \( x = -\frac{\pi}{5} \) is \( -1 \).
| Given function | \( y = \cos^{-1}(\cos x) \) |
| Point of interest | \( x = -\frac{\pi}{5} \) |
| Relevant interval for \( x \) | \( x \in [-\pi, 0] \) (since \( -\pi < -\frac{\pi}{5} < 0 \)) |
| Simplified function in interval | \( y = -x \) |
| Derivative \( \frac{dy}{dx} \) | \( -1 \) |
| Slope at \( x = -\frac{\pi}{5} \) | \( -1 \) |
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Inverse Cosine Function | \( y = \cos^{-1}(u) \) is the inverse of \( u = \cos y \) with range \( [0, \pi] \). |
| \( \cos^{-1}(\cos x) \) | This function simplifies based on the interval of \( x \) to keep the output in \( [0, \pi] \). It is a piecewise linear function. |
| Slope of Tangent | The slope of the tangent line to a curve at a point is given by the value of the derivative of the function at that point. |
| Derivative of -x | The derivative of \( f(x) = -x \) with respect to \( x \) is \( f'(x) = -1 \). |
Additional Information: Graph of y = cos⁻¹(cos x)
The graph of \( y = \cos^{-1}(\cos x) \) is a zigzag pattern, consisting of line segments with slopes alternating between \( 1 \) and \( -1 \).
- For \( x \in [0, \pi] \), the graph is \( y = x \) (slope \( 1 \)).
- For \( x \in [\pi, 2\pi] \), the graph is \( y = 2\pi - x \) (slope \( -1 \)).
- For \( x \in [2\pi, 3\pi] \), the graph is \( y = x - 2\pi \) (slope \( 1 \)).
- For \( x \in [-\pi, 0] \), the graph is \( y = -x \) (slope \( -1 \)).
- For \( x \in [-2\pi, -\pi] \), the graph is \( y = x + 2\pi \) (slope \( 1 \)).
The point \( x = -\frac{\pi}{5} \) lies on the segment \( y = -x \) in the interval \( [-\pi, 0] \). The slope of this segment is constantly \( -1 \).
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