# *HELP* Finding x in [_,_] of Possible Inflection Points

1. ### AnnaSuxCalc

55
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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

### 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. ### HallsofIvy

40,939
Staff Emeritus
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. ### AnnaSuxCalc

55
Uhm ok so...
f''(x) = esinx (½(1+cos2x))-esinxsinx

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

5. ### AnnaSuxCalc

55
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. ### HallsofIvy

40,939
Staff Emeritus
??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. ### AnnaSuxCalc

55
OKAY so:
f'' (x) = esinx(cos2x - sinx)
let cos2x - sinx = 0
1-sin2(x) - sinx = 0
let y = sinx
y2 + y - 1 = 0
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?

### 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. ### AnnaSuxCalc

55
so ok since:
sin(x) = 0.6180
so I drew my sin curve but...
ok so am I supposed to just for example do:
x = 0.6662 + π
??

### 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. ### AnnaSuxCalc

55
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. ### NoMoreExams

626
With a calculator?

13. ### AnnaSuxCalc

55
lol omg *embarrassed* YES

14. ### NoMoreExams

626
Didn't know anyone got so excited about calculus

55

626

17. ### AnnaSuxCalc

55
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. ### NoMoreExams

626
You would use your calculator and type in arcsin(.618)?

19. ### AnnaSuxCalc

55
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. ### NoMoreExams

626
OK let's pick a simpler example. If you had to find sin(x) = 1/2 how would you figure out x?