Answer:
[tex]f(x)=\frac{e^x}{4+e^x}[/tex]
Step-by-step explanation:
a) First derivative:
[tex]f'(x)=\frac{(e^x)'(4+e^x)-e^x(4+e^x)'}{(4+e^x)^2}[/tex]
[tex]f'(x)=\frac{4e^x}{(4+e^x)^2}[/tex]
b) Increasing
f(x) is increasing if f'(x)>0. We can see that (4+e^x)^2 is always positive and 4e^x is also alwayspositive, so f'(x) is always positive. then we have that increasing iterval is [-inf,-inf].
c) and d) NONE f(x) is define on [-inf,inf] and f'(x)>0 for every x, we dont have local minima and max.
e) second derivative:
[tex] f''(x)=\frac{4e^x(4+e^x)^2-8e^x(4+e^x)e^x}{(4+e^x)^4}[/tex]
[tex]f''(x)=\frac{16e^x+4e^{2x}-8e^{2x}}{(4+e^x)^3}[/tex]
[tex]f''(x)=\frac{16e^x-4e^{2x}}{(4+e^x)^3}[/tex]
f) f"(x)>=0 if [tex]4e^x-e^{2x}>=0[/tex] i.e. [tex]x=< ln4][/tex]
Upward on on (-inf, ln4)
f"(x)<0 if [tex]x>ln4[/tex].
Downward on [ln4, +inf)
g) infllections point (ln4, 1/2)
