|
Question 726020: 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 Theo(13342)   (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.
the graph of the equation looks like this:
from this graph, you can see that the zeroes of the graph are around x = -2/3 and x = 1.
from this 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 this 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.
|
|
|
| |
