Homework Help: *HELP* Finding x in [_,_] of Possible Inflection Points

1. Jan 15, 2009

AnnaSuxCalc

The solutions of f''(x) = 0 are the x-coordinates of possible inflection points on the curve y = f(x), where f'' denotes the second derivative of f. The second derivative of f(x) = esinx is:

f'' (x) = esinxcos2x - esinxsinx

Find all x in [0,2π] of possible inflection points.

Uhhhmmmmm...:surprised

also: our hint for this question is:
"keep in mind that the function g(t) = et is never zero for any real number t"
what is going on here?
Thanks in Advance:!!)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 15, 2009

Staff: Mentor

Factor $$e^{sin(x)}$$ from both terms in f''(x) and then set the remaining factor to 0. As a hint in solving the resulting equation cos^2(x) = 1 - sin^2(x).

3. Jan 15, 2009

HallsofIvy

F"(t)= $e^{sin(t)}(cos^2(x)- sin(x))$ and, as your hint tells you, $e^{sin(t)}$ is never 0. In order to have F"(t)= $e^{sin(t)}(cos^2(x)- sin(x))= 0$ you must have $cos^2(x)- sin(x)= 0$.

4. Jan 15, 2009

AnnaSuxCalc

Uhm ok so...
f''(x) = esinx (½(1+cos2x))-esinxsinx

f''(x) =esinx (½ + ½cos2x - sinx)
so what should i do now?

5. Jan 15, 2009

AnnaSuxCalc

so
cos2x = sinx or (½+½cos2x) = sinx ?? is this what you mean?
okay my main problem is that i dont understand what the question is asking for what does the last part of the question mean the 'find all x in [0,2π] of possible inflection points'?
what am I trying to find here?

6. Jan 15, 2009

HallsofIvy

??No one has suggested that you convert $cos^2(x)= (1/2)(1+ cos(2x))$. What has been suggested is that you solve $cos^2(x)- sin(x)= 0$. Mark44 suggested you use $cos^2(x)= 1- sin^2(x)$ so that $cos^2(x)- sin(x)= 0$ becomes $1- sin^2(x)- sin(x)= 0$. That is a quadratic equation in sin(x). If you let y= sin(x) you have $y^2+ y- 1= 0$. Solve that that for y and then solve sin(x)= y for x.

7. Jan 15, 2009

AnnaSuxCalc

OKAY so:
f'' (x) = esinx(cos2x - sinx)
let cos2x - sinx = 0
1-sin2(x) - sinx = 0
let y = sinx
y2 + y - 1 = 0
Quadratic Equation Gives 2 Answers:
0.6180
and
-1.6180 <-- DO NOT use because it is not in the interval in question
sin(x) = y
sin(x) = 0.6180
x = sin-1(0.6180)
x = 38.17
is that it for this problem?

8. Jan 15, 2009

Staff: Mentor

Not quite.
You're correct to discard the value -1.6180, but for the other, you have found one value of x for which sin(x) = .6180. There are two values in the interval of interest, [0, 2pi]. Both values should be in radians, not degrees.

9. Jan 15, 2009

AnnaSuxCalc

so ok since:
sin(x) = 0.6180
x = 0.6662 rad
so I drew my sin curve but...
ok so am I supposed to just for example do:
x = 0.6662 + π
??

10. Jan 15, 2009

Staff: Mentor

No, that's not it. Think about what the graph of y = sin(x) looks like between 0 and 2pi. As a hint, sin(pi/6) = 1/2 (or sin(30 deg.) = 1/2). What's another value x, in radians, for which sin(x) = 1/2? It's not pi + pi/6.

11. Jan 15, 2009

AnnaSuxCalc

oh ok I have to find the where the line
sinx = y = 0.6180 crosses with the curve y = sinx
that is two times in this interval once at the inverse of sin(0.6180) and ???
okay i can see it on my graph but how do i actually calculate this value of x?

12. Jan 15, 2009

NoMoreExams

With a calculator?

13. Jan 15, 2009

AnnaSuxCalc

lol omg *embarrassed* YES

14. Jan 15, 2009

NoMoreExams

Didn't know anyone got so excited about calculus

15. Jan 15, 2009

AnnaSuxCalc

can you help me please

16. Jan 15, 2009

NoMoreExams

Help you with what? I thought your question has been answered

17. Jan 15, 2009

AnnaSuxCalc

No...this question hasnt:
oh ok I have to find the where the line
sinx = y = 0.6180 crosses with the curve y = sinx
that is two times in this interval once at the inverse of sin(0.6180) and ???
okay i can see it on my graph but how do i actually calculate this value of x?

18. Jan 15, 2009

NoMoreExams

You would use your calculator and type in arcsin(.618)?

19. Jan 15, 2009

AnnaSuxCalc

haha, yes I did taht but it only gives me the one value of x, but there are two intersections so there is another value of x that gives sin(x) = 0.6180
if you just do sin-1 (0.6180) = 0.6662 ---> only one value, i need 2 values

20. Jan 15, 2009

NoMoreExams

OK let's pick a simpler example. If you had to find sin(x) = 1/2 how would you figure out x?

21. Jan 15, 2009

AnnaSuxCalc

sin-1(1/2) = x

22. Jan 15, 2009

NoMoreExams

lol sure, but if you didn't have a calculator, how would you figure that one out.

23. Jan 15, 2009

AnnaSuxCalc

oh you got me i dont know

24. Jan 15, 2009

NoMoreExams

You should go back and review trigonometry. It's on the unit circle.

25. Jan 15, 2009

AnnaSuxCalc

come on help me out here (your name), lol ok fine soooo...
sinθ = y/r --> y=1 and r=2 using THE theorem x=sqrt3 ???

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