Question

where is f(x)=ln(7+x^2) concave up and concave down and what are the inflection points

Answer 1

take the first and second derivative... what do you get?

Answer 2

f'(x)= 2x/ (x^2-7) and f"(x)= -2(x^2-7)/ (x^2+7)^2

Answer 3

Concavity is determined by the second derivative.
Concave up = positive second derivative
Concave down = negative second derivative.

Note that f'(x)=(2x)/(7+x^2)
f "(x)=(2(7+x^2)-2x(2x) )/ (7+x^2)^2 = (14-2x^2)/(7+x^2)^2
Now you can continue and solve the inequality f '(x)>0 and f "(x)<0.

Answer 4

but then what is the interval that it is concave up?

Answer 5

-2x^2+14>0, then x<+-sqrt7?

Answer 6

notice f is defined for all reals so you don't have to worry about the domain of f.

Answer 7

potential inflection points are as -sqrt7 and sqrt7
break up your interval using those points.
pick a #, c, from those intervals and stick it into the second derivative.
if f''(c) > 0 , concave up... cc down otherwise.