proyaop

(r, theta) in polar form, where theta indicates the direction and r, the length. Think of (2, pi/3) as a triangle with diagonal two and angle pi/3 = 60 degrees.

Thus, the base of the triangle, is 2/2 = 1, and then the height would be sqrt(3), so Point A has coordinates (1, sqrt(3)).

Point B likewise, in the fourth quadrant, would have the included angle be 30 degrees with a diagonal of 4. Thus, its coordinates would be (2sqrt(3), -2).

Observing, we have 3 overarching fractions in which their denominators cannot be 0, and thus if we set their denominators equal to 0, we can find the 3 values of x that are not in the domain of f(x) (domain means all the values of x that "work/are allowed" for the function):

1) \(1 + {2\over{1 + {3\over{x}}}}=0\)

Solve for x: \(-1 = {2\over{1 + {3\over{x}}}}\); \(-1 - {3\over{x}}=2\); x = -1

2) However, what if \(1 + {3\over{x}} = 0\)? Solving for x, you would get x = -3

3) Lastly, x = 0 is quite obvious, because then 3/0 would be undefined.

Their sum is -1 + (-3) + 0 = -4.

         To not be in the domain of a function, then x must be a value in which the denominator is equal to 0 and thus the expression is undefined. x^2 + 7 can never be 0 for a real value of x, because x^2 is nonnegative. x^3 - x^4 = x^3(1 - x), which this expression is 0 when x = 0 and 1. Additionally, x^2 - 3x + 2 = (x - 2)(x - 1) is 0 for when x = 2 and 1.

         Altogether, we have x = 0, 1, or 2 that will result in one of the fractions having a denominator of 0... Since 1/0 is undefined, x = 0, 1, and 2 are not part of the domain of f.

         Hence, their sum 0 + 1 + 2 = 3, which is your final answer.

If x is 0, then you would get undefined when calculating 1/x, or 1/(x^2), or 1/(x^3), with x = 0 being the only value not in the domain of f(x). Thus, there is ONE real number that is not in the domain of the function above.

\(f(x) = \frac{2x + 3}{x} - 7 + \frac{1}{x - 2} + \frac{x^2}{x + 4}\)

Put this all over a common denominator to get a rational function:

\({(2x + 3)(x - 2)(x + 4)\over(x)(x-2)(x+4)}-{7(x)(x-2)(x+4)\over(x)(x-2)(x+4)}+{1(x)(x+4)\over(x)(x-2)(x+4)}+{x^3(x-2)\over(x)(x-2)(x+4)}\)

Expanding:

\({2x^3+7x^2-10x-24\over{x^3+2x^2-8x}}-{7x^3+14x^2-56x\over{x^3+2x^2-8x}}+{x^2+4x\over{x^3+2x^2-8x}}+{x^4-2x^3\over{x^3+2x^2-8x}}\)

Combine numerators:

\(x^4-7x^3-6x^2+50x-24\over{x^3+2x^2-8x}\)

Clearly, after graphing or observing the degree of the numerator, f will have a range of (-inf, inf).

Substitution: f(k,-3) = 2k - 3(-3)^2 + 7 - 4(k)(-3) + 8(k)^2 = = 2k - 3*9 + 7 + 12k + 8k^2 = 8k^2 + 14k - 20 = -10

8k^2 + 14k - 10 = 0; divide by 2: 4k^2 + 7k - 5 = 0

Use quadratic formula: \(k = {-b \pm \sqrt{b^2-4ac} \over 2a}\); substituting, you have \(k = {-7\pm\sqrt{49+80}\over8}={-7\pm\sqrt{129}\over8}\), two possible values.

Opening parenthesis for only addition never hurts: \(1 - 3i + 2-i+2+i+1+3i\)

One can treat \(i=\sqrt{-1}\) as a variable, and combine like terms normally: \(1+2+2+1-3i+3i-i+i=6\)

Combine like terms: \({15 - 16i\over{-3+4i}}\)

Rationalize the denominator by multiplying by conjugate to get difference of squares which removes the imaginary term:

\({(15 - 16i)(-3 - 4i)\over{(-3 + 4i)(-3-4i)}}={-45+48i-60i+64i^2\over{9-16i^2}}={19-12i\over25}={19\over25}-{12\over25}i\)

Combine like terms: \(x^2 - 4x +48 = 0\)

Quadratic formula: \(x = {4\pm\sqrt{16-4*48}\over2}={4\pm4i\sqrt{11}\over2}=2\pm2\sqrt{11}i\)

So, your answers are \(2 - 2\sqrt{11}i\), \(2 + 2\sqrt{11}i\)