Question #46a96

1 Answer
Dec 9, 2017

Explanation below
graph{6e^(x)-e^(2x) [-14.19, 14.29, -5.05, 9.78]}

Explanation:

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>