Answer the following questions for the function

Question

f(x) = sin^2(x/3)
defined on the interval [ -9.424778, 2.356194].

Rememer that you can enter pi for \pi as part of your answer.

f(x) is concave down on the interval ?

Please help, im REALLY stuck.

Answer 1

our domain is [-3π,3π/4]

f'(x) = 2(sin(x/3))(cos(x/3)
= sin (2x/3)
(I used sin 2A = 2sinAcosA)

f''(x) = (2/3)cos(2x/3)
finding the points of inflection ....
cos(2x/3) = 0
2x/3 = π/2 or 2x/3 = 3π/2
x = 3π/4 or x = 9π/4
but the period of cos(2x/3) = 2π/(2/3) = 3π

so within our domain we would have points of inflection at
x = 3π/4 or appr. 2.356 (the right side end point of domain)
x= 3π/4 - 3π = -9π/4 or appr. -7.1 (inside our domain) or
x = 9π/4 - 3π = -3π/4 or appr. -2.34 (inside our domain)

I use http://rechneronline.de/function-graphs/ to see my graphs.
Enter (sin(x/3))^2 into the function box, and change the "range x-axis" from -10 to 5.
You will see that the above results are correct.

so f(x) is concave down for -9π/4 < x < -3π/4

(remember concave down means f''(x) < 0 , check by picking any x in that spread and calculationg f''(x)
e.g. f''( -5) = (2/3)cos(-10.3) = -.65.. so concave down )

Answer 2

Thanks for posting that link. It's one of my favorite graphing sites.