Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
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Explanation:
Recall that tan(x) = sin(x)/cos(x).
The discontinuities occur whenever cos(x) is equal to zero.
Solving cos(x) = 0 will yield the locations when we have discontinuities.
This all applies to tan(x), but we want to work with tan(x/2) instead.
Simply replace x with x/2 and solve for x like so
cos(x/2) = 0
x/2 = arccos(0)
x/2 = (pi/2) + 2pi*k or x/2 = (-pi/2) + 2pi*k
x = pi + 4pi*k or x = -pi + 4pi*k
Where k is any integer.
If we make a table of some example k values, then we'll find that we could get the following outputs:
- x = -3pi
- x = -pi
- x = pi
- x = 3pi
- x = 5pi
and so on. These are the odd multiples of pi.
So we can effectively condense those x equations into the single equation x = pi+k*pi
That equation is the same as x = (k+1)pi
The graph is below. It shows we have jump discontinuities. These are <u>not</u> removable discontinuities (since we're not removing a single point).
The to that problem is 56
A line passes through the points:
A ( - 4. - 1 )
B ( 1 1/2, 2 )
The point-slope form:
y = m x + b
m = ( 2-(-1)) / (3/2-(- 4)) = ( 2 + 1 ) / 3/2 + 4 ) = 3 / 11/2 = 6 / 11
2 = 6/11 * 3/2 + b
2 = 9/11 + b
b = 2 - 9/11 = 22/11 - 9/11 = 13/11
The equation in the slope-intercept form is:
y = 6/11 x + 13/11 / * 11 ( we will multiply both sides by 11 )
11 y = 6 x + 13
Answer :
A ) - 6 x + 11 y = 13