Function

[Aladdin]

f is the function defined over ]0,+infinity[ by \(\displaystyle f(x)=\frac{lnx}{\sqrt{x}}+1-x\)

Designate by (C) it's curve.

1) a) Determine Limit as x tends to 0 and deduce an asymptote (C).

Limit as x tends to 0 = -infinity . . . So x=0 is a verticle asymptote.

b) Determine limit as x tends to +infinity and show that the straight line (d) of equation y=-x+1 is an asymptote to (C).

Limit as x tends to +inf = -infinity . . .

Lim[f(x)-(-x+1)]=Lim[ln(x)/sqrt{x}] = o as x tends to +infinity so y=-x+1 is an asymptoe to (C).

c)Study the postiton of (C) with respect to (d) .

\sqrt(x) is positive for x belongs to ]0,+inf{.

lnx is positive for x >1 . . .So C above (d) for x>1
lnx is negative for 0<x<1 . . .So (C) below (d) for 0<x<1
lnx = 0 for x=1 . . . So (C) cuts (d) at x=1

d) Show that the derivative of f(x) has the same sign as \(\displaystyle g(x)=-(2x\sqrt{x}-2+lnx)\)

I found the derivative but It has no relation with g(x) . . .

e) Show that 1 is the only root of the equation g(x)=0 and deduce the sign of g(x).
Replace in g(x)=0 . . .
g(x)<0 for 0<x<1
g(x)>0 for x>1 ? ? ?

f) Study the variations of f and draw its table.
Since f'(x) follows g(x) so it has same sence as g(x).

g) Trace (C).

Thanks Again
Help please.

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ \frac{x}{\sqrt x}+1-x, \ Domain \ is \ (0,\infty), \ Range \ is \ (-\infty,1.25], \ vertical \ aysmptote \ ax \ x \ = \ 0.\)

\(\displaystyle \lim_{x\to0^{-1}} f(x) \ is \ undefined, \ \lim_{x\to0^{+}}f(x) \ = \ 1, \ hence \ \lim_{x\to0} f(x) \ doesn't exist.\)

\(\displaystyle See \ graph \ below.\)

[attachment=0:256kklva]pqr.jpg[/attachment:256kklva]

 

[Aladdin]

Sorry Glenn ~ There was a typo in my first post.