Question

how many points of inflection does the function f have on the interval 0≤x≤6 if f"(x)= 2-3sqrt(x)cos^3x ? a)1 b)2 c)3 d)4 e)5

Answer 1

Is this the second derivative of the function?
\[
f''(x)=2-3\sqrt{x}\cos^3(x)
\]

Answer 2

yes

Answer 3

Are you allowed to use graphing calculators?

Answer 4

no and plus i don't own one :/

Answer 5

If we have interval 0 <x< 6 maybe we can create three charts one where we let x <0 like pick x = -1

x = 1 for 0<x<6

and let x = 2pi since it's above 6?

Answer 6

okay can u just show me so i can make sure i understand?

Answer 7

Even Mathematica cannot solve \(f''(x)=0\)... However, it can plot a graph.

Answer 8

Aren't you given a graph along with the question ?

Answer 9

@dolphins121

Answer 10

This is the graph of the function.

Answer 11

I see three of them?

Answer 12

no there is no graph that comes along with it..@ganeshie8

Answer 13

I see an absolute min on the graph

Answer 14

and an absolute max

Answer 15

wait but isn't there a way to solve it without the graph? bcus im not allowed to use a calculator nor am i given a graph..

Answer 16

Okay maybe your teacher wants you do more work.. sketch the graph or figure it out by some other means. Here I found a different version of this q
http://gyazo.com/725b83349d7b30bd49a7d8c2dc379e1a

Answer 17

hmmm we need to sketch manually then.. starting with first derivative and then pick an x less than 0, between 0 to 6, and above 6. positive first derivative means increasing negative first derivative is decreasing... and then when you take the second derivative concave up occurs when your result is positive and concave down is when the result is negative. I remembered having to manually draw those graphs.

Answer 18

wtheck that graph doesn't have inflection and I know what it looks like @ganeshie8

Answer 19

i think thats the graph of f''(x).. idk google gave me that haha

Answer 20

nvm thats the grpah for next question in the problem set :/
looks like we need to graph it manually as you're saying

Answer 21

The function provided is the second derivative. The function itself is not integrable by hand either. Mathematica returned this when integrated f''(x) twice.
\[
\begin{align*}
f(x)&=\frac{1}{48} \left(\sqrt{2 \pi } \left(-81 C\left(\sqrt{\frac{2}{\pi }}
\sqrt{x}\right)+54 x S\left(\sqrt{\frac{2}{\pi }} \sqrt{x}\right)+2
\sqrt{3} x S\left(\sqrt{\frac{6}{\pi }} \sqrt{x}\right)\right)\\-\sqrt{6
\pi } C\left(\sqrt{\frac{6}{\pi }} \sqrt{x}\right)+6 \sqrt{x} \left(8
x^{3/2}+27 \cos (x)+\cos (3 x)\right)\right)
\end{align*}
\]

Answer 22

omg. yo teacher is evil! for assigning that shiz

Answer 23

C is the Fresnel C and S is the Fresnel S.

http://en.wikipedia.org/wiki/Fresnel_integral

Answer 24

ugh it's just so confusing. & haha yeah idk why she assigned a problem this hard :/ @UsukiDoll

Answer 25

I finally got the Mathematica to work for me. x=5.43173 is the x coordinate of the inflection point.

Answer 26

I think you could only use the graph. I don't think there is another way. Maybe Newton-Raphson but I am not familiar with it.

Answer 27

oh. well thanks for trying :) i appreciate it

Answer 28

@ganeshie8 what problem set are you looking at
"nvm thats the graph for next question in the problem set :/
looks like we need to graph it manually as you're saying "

Answer 29

just out of curiosity :)

Answer 30

it might be easier to solve when is f ' ' (x) > 0 , and f ' ' (x) < 0 , you can do this approximately

Answer 31

How do you solve it approximately? I am interested.

Answer 32

I graphed f ' ' (x) using a calculator, and there is a zero between 5 and 6

Answer 33

Since f ' ' is continuous on [0,6]
and since
f ' ' (5) = 1.84688 > 0
f ' ' ( 6) = -4.504916 < 0

By the intermediate value theorem there is some value c between 5 and 6 such that f ' ' (c) = 0.
And since the sign changes there, it is an inflection point

Answer 34

so there is only one inflection point?

Answer 35

yes because f ' ' (x) stays positive between 0 and 5

Answer 36

actually the 'zero' is x = 5.437318

Answer 37

alright well thanks for ur help. I appreciate it:)

Answer 38

I couldn't find this question online to compare it to.

Answer 39

@perl

https://books.google.co.in/books?id=Z4hwkWDh5NIC&pg=PA575&lpg=PA575&dq=%22how+many+points+of+inflection+does+the+function%22%2B%22have+on+the+interval%22&source=bl&ots=pnFB2vVfQ_&sig=qbTLKc9Ff5IdS5DICED_sEW47p4&hl=en&sa=X&ei=ieHNVIiKGcuE8gXmhYC4Cg&ved=0CC4Q6AEwAg#v=onepage&q=%22how%20many%20points%20of%20inflection%20does%20the%20function%22%2B%22have%20on%20the%20interval%22&f=false

Answer 40

thanks @ganeshie8
if you really want to be absolutely there is only one inflection point, you would have to take the derivative of f ' ' (x) , and see where it is increasing/decreasing. ie, you would you have to find f ' ' ' (x)
which i am going to do since i hate loose ends :)

Answer 41

I'm still curious
I googled the phrase "How many points of inflection does the function f have on the interval 0<= x <= 6" and i didnt get that AP calc book. You have some amazing searching skills :)

Answer 42

looks like you searched right into google books