Concavity on intervals

[Violagirl]

Hi, I am unsure of how to solve these two problems. Any help would be greatly appreciated. Thanks!!

1. Determine the open intervals on which the graph is concave upwards or concave dowward.

y = 2x-tan x, [-TT/2, TT/2]

2. Find the points of inflection and discuss the concavity of the graph of the function.

f(x) = sin x/2, [0, 4TT]

 

[BigGlenntheHeavy]

\(\displaystyle 1) \ f(x) \ = \ 2x-tan(x) \ on \ the \ open \ interval \ (-\pi/2,\pi/2).\)

\(\displaystyle f \ ' \ (x) \ = \ 2-sec^{2}(x), \ f \ " \ (x) \ = \ -2sec(x)[sec(x)tan(x)] \ = \ -2sec^{2}(x)tan(x) \ = \ \frac{-2sin(x)}{cos^{3}(x)}\)

\(\displaystyle \frac{-2sin(x)}{cos^{3}(x)} \ = \ 0 \ \implies \ sin(x) \ = \ \frac{0}{-2} \ = \ 0\)

\(\displaystyle Hence, \ x \ = \ arcsin(0) \ = \ 0\)

\(\displaystyle f(0) \ = \ 2(0)-tan(0) \ = \ 0\)

\(\displaystyle Ergo, \ point \ of \ inflection \ is \ (0,0) \ on \ domain \ (-\pi/2,\pi/2).\)

\(\displaystyle (-\pi/2,0), \ f \ " \ (-1) \ > \ 0, \ concave \ up, \ (0,\pi/2), \ f \ " \ (1) \ < \ 0, \ concave \ down\)

\(\displaystyle Do \ 2 \ the \ same \ way.\)