Where on the interval [0,pi/4] is the function f(x)=(7x+3)/(cos(4x)) continuous?
Using interval notation
A good look at the graph might help
http://www.wolframalpha.com/input/?i=plot+f(x)%3D(7x%2B3)%2F(cos(4x))+from+0+to+%CF%80%2F4
Hint: remember we can't divide by zero.
Looks like we have a vertical asymptote when cos(4x) = 0
To determine where the function f(x) = (7x + 3) / cos(4x) is continuous on the interval [0, π/4], we need to analyze two potential points of discontinuity: when the denominator cos(4x) equals zero and when the function is undefined.
1. Points where the denominator is zero:
Since the cosine function is zero at odd multiples of π/2, we set 4x equal to π/2 and solve for x:
4x = π/2
x = π/8
Since π/8 falls within the interval [0, π/4], we need to check this point.
2. Points where the function is undefined:
The function f(x) is undefined when the denominator, cos(4x), equals zero. However, cos(4x) is never equal to zero on the interval [0, π/4] because 4x does not equal π/2 or any odd multiples of π/2 within this interval.
Therefore, the only potential point of discontinuity is x = π/8. To express this in interval notation, we write the continuous portion of the interval as [0, π/8) ∪ (π/8, π/4].