Calculus(I) Problem. find the value of c so that the range is all reals

[eibon]

Homework Statement

For what values of the constant c is the range of F(x) (-infinity,+infinity)

F(x) =g(x)/h(x)

g(x)=x^2 +2x +C
h(x)=x^2 + 4x + 3C

Homework Equations

The Attempt at a Solution

g(x)=(x+1)^2 +(C-1)
h(x)=(x+2)^2+(3C-4)

h(x)>g(x) for any x less then -c
and g(x) has roots at -2 +-(4-3C)^(1/2) and i think that h(-2 -(4-3C)^(1/2)) >0
so F(x) at (-2 -(4-3C)^(1/2) -a very small number) is +infinity
and F(x) at (-2 -(4-3C)^(1/2) +a very small number) is -infinity

so as long as h(x) has two real roots that are not the same then the range for f(x) is all realsso my problem is to show it with some math and not a graphical way

and i think but can't prove very good that c can be any number less then 4/3, if anyone can help or point me in the right direction that would be great

 

[rock.freak667]

and i think but can't prove very good that c can be any number less then 4/3, if anyone can help or point me in the right direction that would be great

Well what I did was divide the polynomials you'd get it in the form


$$F(x)=K +\frac{G(x)}{h(x)}$$

the graph would be discontinuous when the roots of h(x) are real and distinct. This we don't want.

 

[vela]

Well what I did was divide the polynomials you'd get it in the form


$$F(x)=K +\frac{G(x)}{h(x)}$$

the graph would be discontinuous when the roots of h(x) are real and distinct. This we don't want.

How are you going to get it to go to $\pm\infty$ then? F(x) goes to 1 asymptotically in either direction.

 

[Mentallic]

Well what I did was divide the polynomials you'd get it in the form


$$F(x)=K +\frac{G(x)}{h(x)}$$

the graph would be discontinuous when the roots of h(x) are real and distinct. This we don't want.

The question states that the range must be all reals. You're thinking about the domain.

I'm afraid I can't contribute very much to this problem other than while studying the graphs the answer seems to be 0<c<1 because of the following occurences, which hopefully someone else can explain
The idea is that we are looking for the function to switch between $+\infty$ and $-\infty$ around a vertical asymptote and continue through all real y till the next asymptote.

For $c> 4/3$, h(x) has no real roots and the range does not exist for $y\leq 1/4$

$c=4/3$, h(x) has a double root and around the asymptote F(x) is $+\infty$.

$1<c<4/3$ the function between the two distinct asymptotes begins from $-\infty$ and returns back to $-\infty$ without completing the entire range.

$c=1$ we have one of the roots of h(x) being a root of g(x) so it acts like a single asymptote, so we have $y\neq 1$ because of the horizontal asymptote

$0<c<1$ this is the only restriction on c that makes F(x) go from $-\infty$ to $+\infty$ between the two asymptotes. I believe this is the answer.

$c=0$ pretty much the same result as $c=1$

$c<0$ same as $1<c<4/3$

I hope this helps at all. I'd be interested to see some mathematical reasoning to show the answer rather than my half-baked attempt to explain it.