Question

concavity of f(x)=3sin(x)+2(cos(x))^2

Answer 1

Are you allowed to use a graphics utility?

Answer 2

find d2f/dx2.
if it is negative, concave.
if positive, convex
if 0, then both.
try testing your d2f/dx2 within interval [0, pi] for -ve or +ve

Answer 3

no

Answer 4

yes 2nd derivative will tell you the concavity of the function at any given point

f'(x) = 3cos(x) -4sin(x)cos(x)

f''(x) = -3sin(x) -4cos^2(x) +4sin^2 (x)

with a little help from trig identities
--> -3sin(x)-4cos(2x)

Now plug in any x value to determine concavity

Answer 5

wait i think this may be more helpful

f''(x) =-3sin(x) -4cos^2(x) +4sin^2 (x)
= -3sin(x) -4(1-sin^2(x)) +4sin^2(x)
=8sin^2(x) -3sin(x) -4

u = sin(x) find zeros (inflection points)
--> 8u^2-3u-4 = 0
--> u = -.544, .919


concave down:
-.544 < sin(x) < .919
-33 < x < 67

concave up:
-90 < x < -33 and 67<x <90