Answer:
Anything in the form x = pi+k*pi, for any integer k
These are not removable discontinuities.
============================================================
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).
Note that h(x) = 4x^2 - 8x - 60 factors to 4(x^2 - 2x - 15), and that the roots (zeros) of (x^2 - 2x - 15) are -5 and +3. Thus, the smallest zero is -5.
Kindly proofread the question!!!
The leading term is x^6, not x5.
Answer: The additional roots are -2, +2, each with multiplicity 2.
Step-by-step explanation:
x^6 - 16x^2 = 4x^4 - 64
x^6 - 4x^4 - 16x^2 + 64 = 0
(x+2i)(x-2i) = (x^2+4) is a factor.
(x-2)(x+2)(x-2)(x+2)(x^2+4) = 0
Polynomial long division by x^2+4
x^6 - 4x^4 - 16x^2 + 64 = 0
First term of quotient is x^4
Subtract x^6+4x^4
Remainder is -8x^4 -16x^2+64
Second quotient term is -8x^2
Subtract -8x^4-32x^2
Remainder is 16x^2+64
Third quotient term is 16
Subtract 16x^2+64
Remainder is zero
Quotient is x^4 - 8x^2 + 16
(x^2-4)^2 = x^4 + 2(1)(-4)x^2 + 16
(x-2)(x+2)(x-2)(x+2)(x^2+4) = 0