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` .
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