Question

Identify the inflection point for the function f(x) = 8sin x + 2x^2.

Answer 1

Take second derivative and set it equals to 0, that's your inflection point

Answer 2

No whenever the signs change

Answer 3

Like that too and also signs have to change

Answer 4

right it usually happens when f''=0
but we can always choose test numbers before and after each possible inflection point to see if the concavity switches
and if it does then it is an inflection point

Answer 5

so what is f' curiousmath?

Answer 6

(8sinx)'=8(sinx)'=8*?

(2x^2)'=?

Answer 7

fill in the question marks

Answer 8

8cosx and 4x

Answer 9

am i wrong ?

Answer 10

good!

Answer 11

ok so now f'(x)=8cosx+4x
now we need f''
f''(x)=-8cosx+4

Answer 12

now set =0 and solve
oh yeah do you have any restrictions on the domain

Answer 13

okay so now we have to make it equal to 0. and find the three points right?

Answer 14

there is infinitly many possible inflection points

Answer 15

so we have -8cosx+4=0
so -8cosx=-4
cosx=4/8
cosx=1/2
now what x's can you think of that will make this true?

Answer 16

Isn't f''(x) = -8sinx + 4

Answer 17

yes you are right

Answer 18

I feel like ripping my hair out. omg I have no idea!

Answer 19

-8sinx+4=0
-8sinx=-4
sinx=4/8
sinx=1/2

Answer 20

what about when x=pi/6 and 5pi/6
if these are inflection points then so are
x=pi/6+2npi and x=5pi/6+2npi
for n=...,-3,-2,-1,0,1,2,3,....

Answer 21

and if you plug in number before and after pi/6 you see the concavity switches
and if you plug in number before and after 5pi/6 you see the concavity switches
so (pi/6+2npi,f(pi/6+2npi)) and (5pi/6+2npi, f(5pi/6+2npi)) are inflection points

Answer 22

when you pick your test numbers make sure its in the intervals between the possible inflection points and the areas between the possible inflection points and infinity