Please help me I don’t understand $

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  a) f: all real numbers except 1 and 2; g: all real numbers except 2

  b) f(x) = 5/(x -2) . . . . x = 2 is a vertical asymptote

  c) f(x) = -5/4, -5/3, -5/2, undefined, undefined, 5, 5/2

      g(x) = -5/4, -5/3, -5/2, -5, undefined, 5, 5/2

  d) f(x) is also undefined at x=1

Step-by-step explanation:

a) Any question about the domain of a function concerns the values of the variable for which the function is defined, or not. A rational function, such as these, will be undefined when its denominator is zero. So the question of domain becomes a question of the values of x that make the denominator zero.

The zeros of the denominator of f(x) can be found by factoring it.

  x^2 -3x +2 = (x -1)(x -2)

These factors are zero when x=1 and when x=2. Hence, these values are excluded from the domain of f(x).

  domain of f(x): all real numbers except x=1 and x=2

Then, by inspection, the domain of g is similar, but a little different:

  domain of g(x): all real numbers except x=2

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b) Based on our work for part (a), we find that f(x) can be factored and simplified like this.

There will be a vertical asymptote at x=2. The attachment shows a graph of f(x) with its vertical asymptote.*

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c) The difference between the two functions is that f(x) is undefined at x=1.

  x = -2, -1, 0, 1, 2, 3, 4

  f(x) = -5/4, -5/3, -5/2, undef, undef, 5, 5/2

  g(x) = -5/4, -5/3, -5/2, -5, undef, 5, 5/2

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d) The functions differ in that f(x) is undefined at x=1, whereas g(x) is not.

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* The fact that f(x) is undefined at x=1, but there is no vertical asymptote at that point, means the function has a "hole" in its graph there. This is illustrated on the graph by an open circle, meaning the function is not defined at that point.

c. (3,2,6) is the answer.

d. you cannot tell because there are no values shown.