factor x^2 + x - 6 to get (x+3) * (x-1)
equation becomes f(x) = 2 / ((x+3)*(x-2))
when x = -3, the equation becomes undefined.
when x = 2, the equation becomes undefined.
the domain is therefore all real values of x such that x is not equal to -3 and x is not equal to 2.
here's the graph of what the equation looks like.
the graph of the equation is in blue.
the vertical asymptotes are in orange.
the value of y cannot be defined when x = -3 and when x = 2 because the denominator of the equation becomes 0 at those values of x.
g(a) = 1 and h(a) = sqrt(a) + 2a^2 + 1
Find g(h(2)) and g(h(a)).
i believe that the g function will return 1 regardless of what the value of a is.
if you let y = g(a), then you get y = 1.
this function will always return 1 regardless of the value of a.
the graph of this function will be a horizontal line at y = 1.
if h(a) = sqrt(a) + 2 * a^2 + 1, then h(2) will be equal to sqrt(2) + 2 * 2^2 + 1 = sqrt(2) + 9.
you just replace a with 2 and evaluate the function.
h(2) is therefore equal to sqrt(2) + 9.
g(h(2) becomes g(sqrt(2) + 9) which is still equal to 1.
g(a) tells you the argument of the function is the variable called a, but there is no variable called a in the function.
the value of the function will always be equal to 1, regardless of what the argument is.
likewise, g(h(x)) = 1 becomes g(sqrt(x) + 2 * x^2 + 1) = 1 which is also equal to 1 because the value of the function will always be 1 regardless of what the argument is.
that's my take.
hopefully i'm right.
it makes sense to me.
hopefully it makes sense to you and your professor.
let me know if you are successful with this answer and, if not, why.