#f(x)=(x+1)/(sqrt(5x^2+35))#
Vertical asymptote is in the point where it isn't defined that is:
#sqrt(5x^2+35)!=0#
#5x^2+35!=0#
#x^2!=-35/5#
#x!=sqrt(-35/5)#
As you can see this is absolute nonsense. That can't be done. Square root must be #>0#. That means it doesn't have a vertical asymptote and The Domain is #RR#
In order to determine maximum and minimum we need first derivative:
#f(x)=(x+1)/(sqrt(5x^2+35))#
#f^'(x)=((x+1)^'(sqrt(5x^2+35))-(x+1)((5x^2+35)^(1/2))^')/(sqrt(5x^2+35))^2#
#f^'(x)=(1*sqrt(5x^2+35)-(x+1)((1*10x))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2#
#f^'(x)=((2(sqrt(5x^2+35))^2-10x(x+1))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2#
#f^'(x)=(2(5x^2+35)-10x(x+1))/(2(sqrt(5x^2+35))^3#
#f^'(x)=(cancel(10x^2)+2*35cancel(-10x^2)-2*5x)/(2(sqrt(5x^2+35))^3#
#f^'(x)=(cancel2(35-5x))/(cancel2(sqrt(5x^2+35))^3#
#f^'(x)=(5(7-x))/(sqrt(5x^2+35))^3#
#sqrt(5x^2+35)# is always positive so we don't care about that. However #(7-x)# can be + and -
#x in (-oo,7) hArr f^'(x) >0 =># f goes up
#x in (7,oo) hArr f^'(x) <0 =># f goes down
maximum is in x=7#quadf(7)=(7+1)/(sqrt(5*(7)^2+35))=8/(sqrt(280))=4/sqrt70~~0.478#
function doesn't have minimum
horizontal asymptote: #Lim_(xrarr+-oo)f(x)#
1. for #+oo#
#Lim_(xrarroo)(cancelx(1+1/x))/(cancelx(sqrt(5+35/(x^2))))=Lim_(xrarroo)(1+1/x)/(sqrt(5+35/(x^2)))=(1+0)/(sqrt(5+0))=1/sqrt5~~0.447#
- for #-oo#
#Lim_(xrarr-oo)f(x)="by implementing L'Hopitals rule we get"=Lim_(xrarr-oo)(sqrt(5x^2+35)*5x)=-oo#
Intercepts:
#if x=0quad=>quad y=1/sqrt35#
#if y=0quad=>quad 0=x+1quad=>quadx=-1#
