i had to find the domain of f(x)=(x-5)/x^3-13x^2+36x
i came up with (x+9)(x-4)....so i guess x equals 9 or -4.....is this right?
2007-05-20 20:57:24 · 3 answers · asked by mouseblanket 1 in Science & Mathematics ➔ Mathematics
The domain is Real number minus the realnumbers where x^3-13x^2+36x = 0, because when that happens, f is not defined.
x^3-13x^2+36x = 0 solving
x(x-9)(x-4)=0
so x = 0, x = 9 and x = 4 are NOT part of the domain.
2007-05-20 21:06:26 · answer #1 · answered by gjmb1960 7 · 0⤊ 0⤋
no the anwers should be x = -9 and 4 because (x+9)(x-4) must be equal = 0
so if (x+9)(x-4) = 0
x = -9 and 4
2007-05-21 04:01:39 · answer #2 · answered by misshahila 2 · 0⤊ 0⤋
No you check where f'(x)=0
and solve that equation.
The value of x in that equation is the local (maximum and minimum) point of the expression. Compare them and get the global maximum and minimum, the range between them is hte range.
2007-05-21 04:06:58 · answer #3 · answered by TBS 3 · 0⤊ 0⤋