Answer :
To determine the intervals where the function [tex]\( f(x) = \frac{3x + 5}{\cos 2x} \)[/tex] is continuous within the interval [tex]\(\left[0, \frac{\pi}{2} \right]\)[/tex], we need to identify where the function is not continuous.
The function [tex]\( f(x) \)[/tex] will be discontinuous at points where the denominator [tex]\(\cos 2x\)[/tex] is zero because division by zero is undefined.
We know that [tex]\(\cos 2x = 0\)[/tex] at:
[tex]\[ 2x = \frac{(2n+1)\pi}{2} \quad \text{for integers} \quad n \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{(2n+1)\pi}{4} \][/tex]
Within the interval [tex]\(\left[0, \frac{\pi}{2} \right]\)[/tex], we determine the specific values of [tex]\( x \)[/tex] for integer values of [tex]\( n \)[/tex] that fall within this interval.
1. For [tex]\( n = -1 \)[/tex]:
[tex]\[ x = \frac{(2(-1)+1)\pi}{4} = \frac{-\pi}{4} \quad \text{(this value is not within the interval)} \][/tex]
2. For [tex]\( n = 0 \)[/tex]:
[tex]\[ x = \frac{(2(0)+1)\pi}{4} = \frac{\pi}{4} \quad \text{(this value is within the interval)} \][/tex]
3. For [tex]\( n = 1 \)[/tex]:
[tex]\[ x = \frac{(2(1)+1)\pi}{4} = \frac{3\pi}{4} \quad \text{(this value is not within the interval)} \][/tex]
So, within the interval [tex]\(\left[0, \frac{\pi}{2}\right]\)[/tex], the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = \frac{\pi}{4} \)[/tex].
We exclude this point from the interval to determine where the function is continuous. We find that the function is continuous on the following intervals:
[tex]\[ \left[0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right] \][/tex]
In interval notation, the intervals where the function is continuous are:
[tex]\[ \boxed{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right]} \][/tex]
The function [tex]\( f(x) \)[/tex] will be discontinuous at points where the denominator [tex]\(\cos 2x\)[/tex] is zero because division by zero is undefined.
We know that [tex]\(\cos 2x = 0\)[/tex] at:
[tex]\[ 2x = \frac{(2n+1)\pi}{2} \quad \text{for integers} \quad n \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{(2n+1)\pi}{4} \][/tex]
Within the interval [tex]\(\left[0, \frac{\pi}{2} \right]\)[/tex], we determine the specific values of [tex]\( x \)[/tex] for integer values of [tex]\( n \)[/tex] that fall within this interval.
1. For [tex]\( n = -1 \)[/tex]:
[tex]\[ x = \frac{(2(-1)+1)\pi}{4} = \frac{-\pi}{4} \quad \text{(this value is not within the interval)} \][/tex]
2. For [tex]\( n = 0 \)[/tex]:
[tex]\[ x = \frac{(2(0)+1)\pi}{4} = \frac{\pi}{4} \quad \text{(this value is within the interval)} \][/tex]
3. For [tex]\( n = 1 \)[/tex]:
[tex]\[ x = \frac{(2(1)+1)\pi}{4} = \frac{3\pi}{4} \quad \text{(this value is not within the interval)} \][/tex]
So, within the interval [tex]\(\left[0, \frac{\pi}{2}\right]\)[/tex], the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = \frac{\pi}{4} \)[/tex].
We exclude this point from the interval to determine where the function is continuous. We find that the function is continuous on the following intervals:
[tex]\[ \left[0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right] \][/tex]
In interval notation, the intervals where the function is continuous are:
[tex]\[ \boxed{\left[0, \frac{\pi}{4}\right) \cup \left(\frac{\pi}{4}, \frac{\pi}{2}\right]} \][/tex]
