Determine the intervals of continuity of f(x)= ln(sin(x))
I looked at the graph and I have no idea how to state the intervals.
The primary function is ln(...)
we can only take logs of positive numbers
now look at the sine curve only, and note where it lies above the x-axis
so the intervals would be
0 < x < 90° or 360° < x < 450° or 720° < x < 810° or ...
how about something like
360n < x < 360n+90 , for n being an integer
or , in radians
2πn < x < (2πn + π/2) for n ∈ I
To determine the intervals of continuity for the function f(x) = ln(sin(x)), we need to consider the continuity of both the natural logarithm function (ln) and the sine function (sin) separately.
1. The natural logarithm function (ln) is continuous for all positive values. However, it is not defined for non-positive values, including zero. Therefore, ln(sin(x)) is only defined when sin(x) is positive.
2. The sine function (sin) is periodic with period 2π. It takes positive values in the intervals where the graph of sin(x) is above the x-axis.
To find the intervals where sin(x) is positive, we need to look for regions on the unit circle where the y-coordinate of the corresponding point is positive (above the x-axis).
Since sin(x) is positive in the first and second quadrants of the unit circle, we can determine the intervals of continuity for ln(sin(x)) by considering the values of x within each quadrant.
In the first quadrant (0 < x < π/2), sin(x) is positive, so f(x) = ln(sin(x)) is defined and continuous.
In the second quadrant (π/2 < x < π), sin(x) is positive, so f(x) = ln(sin(x)) is defined and continuous.
Therefore, the intervals of continuity for f(x) = ln(sin(x)) are (0, π/2) and (π/2, π).
Remember to always consider the periodicity of trigonometric functions when determining the intervals of continuity.