Graph Sketching-----The hardest one in A-LEVEL

f(x)= -x + Sqrt[ x^3/(x+2)] for x<-2 or x>0

ai) for x>0

f(x)= -x+x Sqrt[ x/(x+2)] = -x + Sqrt[ x^3 /(x+2) ]
f'(x) = -1 +(x+3)Sqrt[ x/(x+2)^3]

f''(x) = 3x/ Sqrt[ x^3 (x+2)^5]

aii) if x<-2

f(x)= -x-xSqrt[x/(x+2)]=-x+Sqrt[x^3/(x+2)]; when absorb the -x into the root, we change to +. since x<-2
f'(x) = -1 +(x+3)Sqrt[x /(x+2)^3]
f''(x) = 3x/ Sqrt[ x^3 (x+2)^5]

bi) f'(x)>0

since f''(x) >0 for x>0 thus, f'(x) is increasing
f'(0)= -1 <0
limit x->inf f'(x) = 0
therefore, all f'(x) <0 for x>0
thus, no solution for x>0

for x<-2
f''(x) >0 => f'(x) increasing
f'(-3)=-1
therefore, the root should large than -1
f'(x)=0
Sqrt[x] (x+3)>Sqrt[ (x+2)^3]
x (x+3)^2>(x+2)^3
x>-8/3

bii) for x>0
f''(x)>0 so, the solution is x>0

for x<-2
f''(x)>0, the solution is x<-2

c)
the relative extreme occur when
f'(x) =0 and f''(x) !=0
or
at the boundary of domain,

since f'(x) occur at x=-8/3 and f''(8/3)>0, which give 1 relative extreme at x=-8/3, which is minimum.

and f''(x) >0 for x<-2,

there are 2 more relative extreme, one is x=-inf, one is x=-2, which are maximum

another relative extreme is at the boundary,
since f'(x) <0 for x>0, i.e f(x) is falling for x>0, i.e. f(0)=0 is the local extreme, which is maximum

d)asymptote when x<<-2

f(x) = -x + Sqrt[ x^3/(x+2)] = -x + Sqrt[ x^2 / ( 1+2/x ) ] ~ -x + Sqrt[ x^2] ~ -x-x= -2x

x>>0

f(x) = -x + Sqrt[ x^3/(x+2)] = -x + Sqrt[ x^2 / ( 1+2/x ) ] ~ -x + Sqrt[ x^2] ~ -x+x =0
but this is not asymptote, since f(0) =0 and f'(x) <0 for x>0

e) i cannot sketch.... but i discripe.

from -inf, f(x)~ -2x, and f(-8/3) is the minimum point, than goes up to +inf at x=-2

from x=0, f(0)=0, f'(0) = -1 and goes down slowly since 0 >f'(x) > -1 for x>0 at +inf, f'(x)=0 so f(x) will become horizontal when x increase