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

[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) = e^sinx is:

f'' (x) = e^sinxcos²x - e^sinxsinx

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

Uhhhmmmmm...

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:!)

 

[Mark44]

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).

 

[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 $cos^2(x)- sin(x)= 0$.

 

[AnnaSuxCalc]

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).

Uhm ok so...
f''(x) = e^sinx (½(1+cos2x))-e^sinxsinx

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

 

[AnnaSuxCalc]

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 $cos^2(x)- sin(x)= 0$.

so
cos²x = sinx or (½+½cos2x) = sinx ?? is this what you mean?
okay my main problem is that i don't 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?

 

[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.

 

[AnnaSuxCalc]

OKAY so:
f'' (x) = e^sinx(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?

 

[Mark44]

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.

 

[AnnaSuxCalc]

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.

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
??

 

[Mark44]

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.

 

[AnnaSuxCalc]

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.

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?

 

[NoMoreExams]

With a calculator?

 

[AnnaSuxCalc]

With a calculator?

lol omg *embarrassed* YES

 

[NoMoreExams]

Didn't know anyone got so excited about calculus

 

[AnnaSuxCalc]

Didn't know anyone got so excited about calculus

can you help me please

 

[NoMoreExams]

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

 

[AnnaSuxCalc]

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

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?

 

[NoMoreExams]

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

 

[AnnaSuxCalc]

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

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

 

[NoMoreExams]

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

 

[AnnaSuxCalc]

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

sin^-1(1/2) = x

 

[NoMoreExams]

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

 

[AnnaSuxCalc]

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

oh you got me i don't know

 

[NoMoreExams]

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

 

[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 ?

 

[NoMoreExams]

You can look the answer up on wikipedia, it's a common angle and it was already addressed in this thread I believe.

 

[AnnaSuxCalc]

okay I am not sure what your getting at, I know that sin(π/6) = 1/2 and sin(5π/6) = 1/2

 

[NoMoreExams]

I'm trying to get you to remember how those values were determined. Sin(pi/6) = 1/2 is easy. So we know 30 degrees is what we need. Since we have sine, it's positive in 1st and 2nd quadrants. So in the first quadrant we have 30 degrees, in the 2nd we know our "line" must make a 30 degree angle with the horizontal which would be pi, so we did pi - pi/6 = 5pi/6

 

[AnnaSuxCalc]

oh ok so my number is not 1/2 but 0.6180 so that is 38.17degrees
so the line must make a 38.2degree angle with the horizontal since pi is (180degrees)
180-38.2 = 141.2degrees ~2.46rad
is this the answer?

 

[NoMoreExams]

Well try plugging that into your calculator :)