Thread: Still lost part 11

Any help would be greatly appreciated.

Thanks once again.
Luis

ok I read the attached...but i could not understand what you want to know...
what do you need help with???

or did you just want someone to check your work??

dan

Dan,

This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

Thanks,
Luis

Originally Posted by FLO21I99

Dan,

This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

Thanks,
Luis

I know exactly what you mean, but NEVER admit that when teachers are around.

-Dan

Originally Posted by topsquark

I know exactly what you mean, but NEVER admit that when teachers are around.

-Dan

haha... we should start a club....

dan

Originally Posted by FLO21I99

Dan,

This is an online class I am taking; (which I would not recommend to anyone, take the class in a traditional setting) check my work, give me the anwers, it doesn't really matter. These are the only nine questions I can't answer. I just want to get them done. Call me lazy, whatever...some things just weren't meant for people to understand and I think algebra is it for me.

Thanks,
Luis

In the first question shown below in the attachment, we see that g(x)=a x^3
grows more quickly as x -> +inf, than does f(x)=x^3.

So choose lets say x=2, then we see that:

8 a>8,

so a>1.

RonL

Originally Posted by FLO21I99

Any help would be greatly appreciated.

Thanks once again.
Luis

Second question (shown in attachment).

If you multiply out the factors the leading lerm is x^3. This dominates
when |x| is large. So as x-> -inf f(x) ->-inf, and as x->inf f(x)->inf.

Also f(x)=0 has three roots: 4, 0, -1. So

f(x)<0, when x<-1,
f(x)>0, when -1<x<0,
f(x)<0, when 0<x<4,
f(x)>0, when x>4

So f(x)>0 on (-1,0)U(4,inf)

RonL