What is the largest interval on which the function is concave down for #f(x) = 5sin(x) + (sin(x))^2# over the interval #[-pi/3, 2pi/3]#?

Question

Luke Alvarez · Accepted Answer

#(0, (2pi)/3]#, open on the left and closed on the right.

f(x) is periodic with period #2pi#. The half of a full wave, in #(0, pi)#, is up, up to 8 units at #x = pi/2#, while the shorter half, in #(pi, 2pi)# goes down up to -4 units at #x = (3pi)/2#. The tangent crosses the wave at #x=0, pi, 2pi#, to turn the curve from concavity to convexity, and vice versa. Of course, viewing from the x-axis, it is looking concave, both sides. I think you would say OK to my answer now..

Charles Arocha · Answer

If you need exact values, I got #[arcsin((-5+sqrt57)/8),(2pi)/3]#.

Which is about #[0.3244, (2pi)/3]#.

#f'(x) = 5cosx+2sinxcosx = 5cosx+sin(2x)# #f''(x) = -5sinx+2cos(2x) = -5sinx+1(1-2sin^2x)# #f''(x) = -4sin^2x-5sinx+2# Solving #f''(x) = 0# we find the partition numbers using #sinx = (-5+sqrt57)/8# The only value of #x# in the interval with this sine is #arcsin((-5+sqrt57)/8) ~~0.3244# For #-pi/3 < x < 0.3244#, we have #f''(x) > 0#, so #f# is concave up. (Use #x=0# as a test number.) For # 0.3244 < x < (2pi)/3#, we have #f''(x) < 0#, so #f# is concave down. (Use #x=pi/2# as a test number.) So the largest interval is the closed interval #[arcsin((-5+sqrt57)/8),(2pi)/3]#. Which is about #[0.3244, (2pi)/3]#.

Aurora Alvarado · Answer

undefined To find where the function $ f(x) = 5\sin(x) + \sin^2(x) $ is concave down, you need to determine where its second derivative is negative. First, find the second derivative of $ f(x) $, then solve for where it is negative over the given interval $[- \frac{\pi}{3}, \frac{2\pi}{3}]$.