# consider the function f(x)= (e^x)/(4+(e^x)) A) Then f'(x) = B) The interval of increase for f(x) C) The interval of decrease for f(x) D) f(x)  has a local minimum at E) f(x) has a local maximum at F) Then = G) The interval of upward concavity for f(x) is H) The interval of downward concavity for f(x) is I) f(x) has an inflection value, x = ?

A) You need to find derivative of function using quotient law such that:

`f'(x) = ((e^x)'*(4+(e^x)) - (e^x)*(4+(e^x))')/((4+(e^x))^2)`

`f'(x) = ((e^x)*(4+(e^x)) - (e^x)*(e^x))/((4+(e^x))^2)`

`f'(x) = (4e^x + e^(2x) - e^(2x))/((4+(e^x))^2)`

Reducing like terms yields:

`f'(x) = (4e^x)/((4+(e^x))^2)`

B) Notice that `4e^x gt 0`  and `((4+(e^x))^2) gt 0` , hence f'(x)>0 for `x in (-oo,oo), ` hence, the function increases over R set.

C) The function only increases over R set.

D) Since `f'(x)!=0`  for any `x in R`  set, hence the function has no local minimum.

E) Since `f'(x)!=0`  for any `x in R`  set, hence the function has no local maximum either.

G) You need to evaluate f''(x) to find where the function is concave up or concave down such that:

`f''(x) = ((4e^x)'*((4+(e^x))^2) - (4e^x)*((4+(e^x))^2)')/((4+(e^x))^4)`

`f''(x) = (4e^x*((4+(e^x))^2) - 2e^x(4e^x)*((4+(e^x)))/((4+(e^x))^4)`

You need to factor out `(4e^x)*((4+(e^x))`  such that:

`f''(x) = (4e^x)*((4+(e^x))(4 + e^x - 2e^x)/((4+(e^x))^4)`

`f''(x) = (4e^x)*(4- e^x)/((4+(e^x))^3)`

You need to solve f''(x) = 0 such that:

`(4e^x)*(4 - e^x)/((4+(e^x))^3) = 0 =gt (4e^x)*(4 - e^x) = 0`

Since `4e^x gt 0 =gt 4 - e^x = 0 =gt e^x = 4 =gt x = ln 4`

If `x lt ln 4 =gt f''(x) gt 0` , hence the function is concave up over `(-oo,ln 4).`

H) Notice that for `x gt ln 4 =gt f''(x) lt 0` , hence the function is concave down over `(ln 4, oo).`

I) Notice that the function has an inflection point at`x = ln 4` .