Question

-9sin^2(x) cos(x) is concave up on what interval?

*function is already in second derivative form

Answer 1

again???

Answer 2

last time you just helped me with concave down

Answer 3

and i havent been able to figure concave up since

Answer 4

hold the phone, it is either concave up or concave down. we found it to be concave up where this is positive right? as i recall it was on the interval where cosine is negative, namely \((\frac{\pi}{2},\frac{3\pi}{2})\)

Answer 5

wait, i never saw (pi/2, 3pi/2) srry

can you help me with 1 to 2 more questions pls

Answer 6

sure if they are quick because i am about to turn in to a pumpkin

Answer 7

i need to know the x-coordinate of the point of inflection of this

Answer 8

-9sin^2(x) cos(x)

Answer 9

that is where the second derivative changes sign. there are two of them, at \(x=\frac{\pi}{2}\) and also at \(x=\frac{3\pi}{2}\)

Answer 10

oh ok this is the original equation of the function

3cosx−cos^(3)x for 0<x<2π

can you let me know x-intercept and local maximum

Answer 11

pls

Answer 12

*original function of the derivative

Answer 13

we already had the derivative, i believe it was \(-3\sin^3(x)\) right?

Answer 14

yeah

Answer 15

this changes sign when \(\sin(x)\) does, at \(x=0\) and \(x=\pi\)

Answer 16

since \(x=0\) is the endpoint of the interval, you only have one local max at \(x=\pi\)

Answer 17

scratch that, \(x=\pi\) is a local minimum, not a local max sorry

Answer 18

yea

Answer 19

to summerize, your funciton is decreasing on \((0, \pi)\) and increasing on \((\pi,2\pi)\) local min at \(x=\pi\)

Answer 20

and we already took care of concave up and down etc

Answer 21

yeah

Answer 22

would that make my local max 2pi?