SOLUTION: Let f(x)=2x^2+7x-4/x^3-1 A. What is the domain of f(x)? B.What is the range of f(x)? C.What are the roots of f(x)? D.determine the end behavior of f(x)? E.Where will

Algebra ->  Probability-and-statistics -> SOLUTION: Let f(x)=2x^2+7x-4/x^3-1 A. What is the domain of f(x)? B.What is the range of f(x)? C.What are the roots of f(x)? D.determine the end behavior of f(x)? E.Where will      Log On


   

Question 935794: Let f(x)=2x^2+7x-4/x^3-1
A. What is the domain of f(x)?
B.What is the range of f(x)?
C.What are the roots of f(x)?
D.determine the end behavior of f(x)?
E.Where will f(x) intersect the y-axis?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
You wrote 2x^2+7x-4/x^3-1 =2x%5E2%2B7x-4%2Fx%5E3-1 ,
but I suspect that you meant f(x)=(2x^2+7x-4)/(x^3-1)=%282x%5E2%2B7x-4%29%2F%28x%5E3-1%29 .
If f%28x%29=%282x%5E2%2B7x-4%29%2F%28x%5E3-1%29 ,
A. the domain of f%28x%29 is all the values of x that give a value to f%28x%29 .
That means all the values of x that
make x%5E3-1%3C%3E0 , and that is all the values of x such that
x%3C%3E1 ,
because x%5E3-1=0--->x%5E3=1--->x=1 .
Other that for x=1 , f%28x%29 is a continuous function, so it varies continuously.
It is zero and it changes sign at the points where 2x%5E2%2B7x-4=0 .
It has no absolute maximum or minimum, because
as x approaches 1 ,
the numerator approaches 2%2A1%5E2%2B7%2A1-4=2%2B7-4=5 ,
while the denominator approaches 0,
which makes the absolute value of f%28x%29 to increase without bounds.
At x=1 the denominator changes sign,
with x%3C1<--->x%5E3-1%3C0 and x%3E1<--->x%5E3-1%3E0 ,
so f%28x%29--->-infinity as x approaches 1 from the left, and f%28x%29--->%2Binfinity as x approaches 1 from the right.

B. The range of f%28x%29 is all the values f%28x%29 can take.
That is all real numbers or %22%28%22-infinity%22%2C%22infinity%22%29%22 .
Graphing the function would show you that,
but I am not sure how you are expected to demonstrate that.
One way would be to demonstrate that
%282x%5E2%2B7x-4%29%2F%28x%5E3-1%29=k<--->2x%5E2%2B7x-4=k%28x%5E3-1%29<--->kx%5E3-2x%5E2-7x-%284%2Bx%29=0%29 has a solution for every possible real value of k .
It is obvious that it is so,
because kx%5E3-2x%5E2-7x-%284%2Bx%29 is an odd-degree polynomial,
an those always hit zero somewhere.

C. The roots of f%28x%29 are the roots of 2x%5E2%2B7x-4 ,
and we find them by solving 2x%5E2%2B7x-4=0 .
Factoring we get
2x%5E2%2B7x-4=0-->2x%5E2-x%2B8x-4=0-->x%282x-1%29%2B4%282x-1%29=0-->%28x%2B4%29%282x-1%29=0 .
So, the roots are x=-4 and x=1%2F2 .

D. The end behavior of f%28x%29 is given by
lim%28+x-%3Einfinity%2C+%282x%5E2%2B7x-4%29%2F%28x%5E3-1%29+%29+=0 .
The limit is so because rational functions tend to zero at the ends
when the degree of the denominator (3 in this case)
is greater than the degree of the denominator (2 in this case).
So, y=0 (the x- axis) is the horizontal asymptote.
The function is positive for x%3E1%3E1%2F2%3E-4 ,
so it approaches the positive x-axis from above.
At the other end, where x%3C-4%3C1%2F2%3C1 ,
f%28x%29%3C0 approaches the negative x-axis from below.

E.Since the y-axis is the line x=o ,
f(x) intersect the y-axis at the point with y=f%280%29=4 , (0,4).