Determine all local extrema, concavity and inflection point/s for y = -2cos(x) in (0, 2Pi). Test by both First Derivative Test and Second Derivative Test and sketch the graph.
To determine the local extrema, concavity, and inflection points of the function \( y = -2\cos(x) \) in the interval \( (0, 2\pi) \), we'll need to take the first and second derivatives of the function.
1. First Derivative:
Find the first derivative of \( y \) with respect to \( x \) by applying the chain rule:
\( y' = \frac{d}{dx}(-2\cos(x)) \)
Differentiating, we get:
\( y' = 2\sin(x) \)
2. Local Extrema:
To find the local extrema, set the first derivative equal to zero and solve for \( x \):
\( 2\sin(x) = 0 \)
Solving this equation results in \( x = 0, \pi, 2\pi \).
We need to check if these points are within the interval \( (0, 2\pi) \) to consider them as local extrema.
\( x = 0 \) is within the interval.
\( x = \pi \) is within the interval.
\( x = 2\pi \) is not within the interval, so we exclude it.
So, the potential local extrema are \( x = 0 \) and \( x = \pi \).
To determine if these are minimum or maximum points, we can use the First Derivative Test:
- Plug in a value less than 0 into the first derivative to the left of \( x = 0 \). For example, \( x = -\frac{\pi}{4} \).
\( y'(-\frac{\pi}{4}) = 2\sin(-\frac{\pi}{4}) = -\sqrt{2} \)
Since the derivative is negative, this suggests a local maximum.
- Plug in a value between 0 and \( \pi \) into the first derivative. For example, \( x = \frac{\pi}{4} \):
\( y'(\frac{\pi}{4}) = 2\sin(\frac{\pi}{4}) = \sqrt{2} \)
Since the derivative is positive, this suggests a local minimum.
Therefore, at \( x = 0 \), we have a local maximum, and at \( x = \pi \), we have a local minimum.
3. Second Derivative and Concavity:
Now, let's find the second derivative of \( y \) with respect to \( x \) to determine the concavity:
\( y'' = \frac{d}{dx}(2\sin(x)) \)
Differentiating, we get:
\( y'' = 2\cos(x) \)
For concavity, we need to find where the second derivative equals zero or is undefined.
\( 2\cos(x) = 0 \)
Solving the equation, we find \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).
Again, we check if these points are within the interval \( (0, 2\pi) \).
\( x = \frac{\pi}{2} \) is within the interval.
\( x = \frac{3\pi}{2} \) is within the interval.
So, the potential points of inflection are \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \).
4. Sketching the Graph:
- The graph of \( y = -2\cos(x) \) has a maximum at \( x = 0 \) and a minimum at \( x = \pi \). These are the local extrema.
- The function changes from concave up to concave down at \( x = \frac{\pi}{2} \), indicating a point of inflection.
- The function changes from concave down to concave up at \( x = \frac{3\pi}{2} \), indicating another point of inflection.
By using the First and Second Derivative Tests, we have identified the local extrema and inflection points, and determined the concavity of the function \( y = -2\cos(x) \) in the interval \( (0, 2\pi) \).