Question

F(x) = 3 sin x + 3 cos x

0 ≤ x ≤ 2π. Find the interval in which f is concave up and concave down. Find the inflection points

Answer 1

F(x) = 3 sin x + 3 cos x
F'(x) = 3cosx-3sinx

to find inflection points, just take the second derivative, set it to zero and solve for x.
so the second derivative is -3sinx-3cos x
you'll have to take 2 conditions. 1) f''(x) = 0 2) f''(x) = UND depends on the function of course when you set your derived function to zero, you'll get numbers, those are your x's and to find the y's to finish up your process , plug in the x you found in the original function, and with that, you'l have your inflection points

Answer 2

f ( x ) = 3 sim x + 3 cos x
f ` ( x ) = 3 cos x - 3 sin x
f `` ( x ) = - 3 sin x - 3 cos x = - 3 ( sin x + cos x )
The inflection points:
- 3 ( sin x + cos x ) = 0
sin x + cos x = 0
sin x = - cos x / : cos x
tan x = - 1
x 1 = 3π / 4
x 2 = 7π / 4
The function is concave up when f``(x) > 0
- 3( sin x+ cos x ) > 0
sin x + cos x < 0
tan x < - 1
f is concave up for:
x ∈ ( π/2, 3π/4 ) ∪ ( 3π/2, 7π/4)
f is concave down for:
x ∈ ( 0, π/2 ) ∪ ( 3π / 4, 3π/2 ) ∪ ( 7π / 4, 2 π ).