Bobosharif S. answered • 01/15/18
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Mathematics/Statistics Tutor
1 Asymptes:
a) Vertical: x=-3, because
Limx→-3(x3/(3+x)2) = - ∞. There is no more vertical asymptotes.
b) Horizontal
Limx→-∞(x3/(3+x)2) = - ∞
Limx→∞(x3/(3+x)2) = + ∞
Therefore, there are no horizontal asymptotes.
c) Slant (oblique) asymptote is of the form y=ax+b, where
a=limx→±∞ (f(x)/x)
and
b=Limx→±∞(f(x)-ax)
Now
a=limx→±∞ (f(x)/x)= limx→±∞ (x3/(3+x)2/x)=1
b=Limx→±∞(f(x)-x)=Limx→±∞(x3/(3+x)2-x)=-6.
So y=x-b is the oblique asymptote.
The function is monotonic. To find (monotonic) increasing (decreasing)
regions,you can use derivatives.
The function increases in (-∞,-9)∪(-3, +∞) and decreases in (-9, -3).
If some questions, feel free to ask.
Bobosharif S.
Yes. The task is to find (actually it is known now) where the function increases, where decreasing. etc. How to do find that? One of the ways is using derivative(s). If you are familiar with derivative, then take derivative of the function:
f(x)=x3/(3+x)2
and then solve for x equation f'(x)=0. Roots of this equation gives you extreme points (including max and min). Once you have these points, you have to find out are those pints are maximum, minumum or neither. By finding those points basically you separated the whole domain of the function into intervals, where f(x) behaves differently. It seems to me you had to know these things, and it is a little but too much to write everything.
I stop here, if any further questions, feel fee to write me. (Even I can explain you somehow online, if you wish).
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01/15/18
Cruel W.
01/15/18