f(x)=e^x[6-e^x]
D(f)=RR
Doesn't have a vertical asymptote
f(-x)=e^-x[6-e^-x]!=f(x)!=-f(x)
This function isn't odd and neither even
isn't periodic function (han no sinx, cosx...)
Inceptions:
If x=0 => y=e^0[6-e^0]=5 .......[0,5]
If y=0 => 0=e^x[6-e^x]
e^x>0 so we have to decide for (6-e^x)
6-e^x=0 =>6=e^x
ln6=x=1.792.................................[ln6,0]
f^'(x)=[6e^x-e^(2x)]^'
f^'(x)=6e^x-2e^(2x)=2e^x(3-e^x)
2e^x>0 so we decide for (3-e^x)
3-e^x=0
3=e^x
ln 3=x=1.098...
x in (-oo,ln3)hArr f^'(x)>0 => f uarr
x=ln3 => "maximum"
x in (ln3,oo)hArr f^'(x)<0 => f darr
f^''=[6e^x-2e^(2x)]^'
f^''=6e^x-4e^(2x)
f^''=2e^x(3-2e^x)
again 2e^x>0 so we decide for (3-2e^x)
3-2e^x=0
3/2=e^x
ln (3/2)=x=0.4054...
x in (-oo,ln(3/2))hArr f^''(x)>0 => "f(x) is convex" uu
x in (ln(3/2),oo)hArr f^''(x)<0 => "f(x) is concave" nn
Behaving in the +oo
Lim_(xrarroo)e^x[6-e^x]=oo*(-oo)=-oo
Behaving in the -oo
Lim_(xrarr-oo)e^x[6-e^x]=0*(6-0)=0
The range: Since maximum is in the point x=ln3 we can calculate that:
f(ln3)=e^ln3[6-e^(ln3)]=3[6-3]=9
H(f)=(-oo,9>