SOLUTION: Give (a) the domain and (b) the zeros of f. f(x)=3x^3-x^2-2x/x(x+1)^2 I would like to thank you for sparing your time to help me and if it is possible, send a response ASAP .

Question 725787: Give (a) the domain and (b) the zeros of f.
f(x)=3x^3-x^2-2x/x(x+1)^2
I would like to thank you for sparing your time to help me and if it is possible, send a response ASAP .... Thank You very much.
Answer by General_Lee87(58) (Show Source): You can put this solution on YOUR website!
The domain appears to be all values of x except x = -1 and x = 0
The range appears to be all values of y.
The zeroes (roots) appear to be x = -2/3 and x = 1.
you find the domain by looking for all values of x that will result in a value of y.

all values of x will result in a value of y except when x = -1 and when x = 0.
when x = -1 or when x = 0, the denominator of the original equation will be equal to 0 which makes the value of y undefined.

therefore x = -1 and x = 0 can't be in the domain.
since there are no further restrictions as to what the value of x can be, then the domain is all values of x except at x = -1 and x = 0.

the zeroes of the equation are found by setting the equation equal to 0 and solving for x.
it's best to simplify the equation first.
start with (3x^3 - x^2 - 2x) / (x * (x+1)^2
since (3x^3 - x^2 - 2x) / x is equal to (3x^2 - x - 2), the equation becomes:
(3x^2 - x - 2) / (x+1)^2
set this equal to 0 to get:
(3x^2 - x - 2) / (x + 1)^2 = 0
multiply both sides of the equation by (x + 1)^2 to get:
(3x^2 - x - 2) = 0
factor to get:
(3x + 2) (x - 1) = 0
solve for x to get:
x = -2/3 or x = 1
these are the roots of the quadratic equation which is the point at which the equation crosses the x-axis which is the zeroes of the equation.

you can confirm that the solutions are correct by graphing the equation.
from the graph, you can see that the zeroes of the graph are around x = -2/3 and x = 1.

from the graph, you can see that there appears to be an asymptote at around x = -1 although it's not real easy to see that.

from the graph, you cannot see that there is a hole at x = 0
the value of y is undefined at x = -1 and also undefined at x = 0.
this is because the denominator of the equation is 0 at those values of x.
at x = -1, this results in an asymptote.
at x = 0, this results in a hole.
if you try to find the value of y when x = 0 and when x = -1, you will not be able to.

you can find a value of y for any other value of x other than 0 or -1.

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