State the intervals where the graph of y = sec(2x) is increasing in between
- pi < theta < pi
The Correct Answer and Explanation is :
To determine the intervals where the graph of ( y = \sec(2x) ) is increasing, we need to follow a few steps. First, we should find the derivative of the function, as the derivative tells us the slope of the graph. If the derivative is positive, the function is increasing.
1. Finding the Derivative:
We begin by taking the derivative of ( y = \sec(2x) ) with respect to ( x ). To differentiate ( \sec(2x) ), we use the chain rule. The derivative of ( \sec(u) ) is ( \sec(u) \tan(u) ), and then we multiply by the derivative of the inside function, which is ( 2x ), giving us:
[
\frac{d}{dx}[\sec(2x)] = 2 \sec(2x) \tan(2x)
]
2. When is the Derivative Positive?
The graph of the function is increasing where the derivative is positive, which means:
[
2 \sec(2x) \tan(2x) > 0
]
Since ( \sec(2x) ) is always positive when ( \tan(2x) ) is positive, we need to determine where ( \tan(2x) ) is positive. The tangent function is positive in the first and third quadrants. Therefore, we need to find the intervals where ( \tan(2x) > 0 ).
3. Solving ( \tan(2x) > 0 ):
We know that the tangent function has a period of ( \pi ). Hence, ( \tan(2x) ) has a period of ( \frac{\pi}{2} ). For ( \tan(2x) ) to be positive, ( 2x ) must be in the first or third quadrant, i.e., in the interval ( (0, \pi) ) and ( (\pi, 2\pi) ), which gives:
[
0 < 2x < \pi \quad \text{or} \quad \pi < 2x < 2\pi
]
Dividing by 2, we get the intervals where ( x ) is increasing:
[
0 < x < \frac{\pi}{2} \quad \text{or} \quad \pi < x < \frac{3\pi}{2}
]
4. Conclusion:
The intervals where the graph of ( y = \sec(2x) ) is increasing are:
[
(0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2})
]
These are the intervals where the derivative is positive, meaning the function is increasing.