Consider the following.
1 answer:
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).
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Answer:
d = 5
Step-by-step explanation:
We want to know what d would be when x = -1, so we can start off by plugging in -1 for x to get the following equation: . On the left hand side, 2 negatives become a positive sign, so the -(-1) becomes a +1, and 8+1 would equal 9. On the right hand side, we just multiply 2 and -1 to get -2. So, the equation now looks like this:
. We know that
= 3, so we would get 3 = d-2. THen we would add 2 to both sides to get d = 5. Hope this helps!