Concavity on intervals

Violagirl

Junior Member
Joined
Mar 9, 2008
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87
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]
 
1) f(x) = 2xtan(x) on the open interval (π/2,π/2).\displaystyle 1) \ f(x) \ = \ 2x-tan(x) \ on \ the \ open \ interval \ (-\pi/2,\pi/2).

f  (x) = 2sec2(x), f " (x) = 2sec(x)[sec(x)tan(x)] = 2sec2(x)tan(x) = 2sin(x)cos3(x)\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)}

2sin(x)cos3(x) = 0      sin(x) = 02 = 0\displaystyle \frac{-2sin(x)}{cos^{3}(x)} \ = \ 0 \ \implies \ sin(x) \ = \ \frac{0}{-2} \ = \ 0

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

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

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

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

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