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^{sinx}cos^{2}x - e^{sinx}sinx 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
Factor [tex]e^{sin(x)}[/tex] 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).
F"(t)= [itex]e^{sin(t)}(cos^2(x)- sin(x))[/itex] and, as your hint tells you, [itex]e^{sin(t)}[/itex] is never 0. In order to have F"(t)= [itex]e^{sin(t)}(cos^2(x)- sin(x))= 0[/itex] you must have [itex]cos^2(x)- sin(x)= 0[/itex].
Uhm ok so... f''(x) = e^{sinx} (½(1+cos2x))-e^{sinx}sinx f''(x) =e^{sinx} (½ + ½cos2x - sinx) so what should i do now?
so cos^{2}x = 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?
??No one has suggested that you convert [itex]cos^2(x)= (1/2)(1+ cos(2x))[/itex]. What has been suggested is that you solve [itex]cos^2(x)- sin(x)= 0[/itex]. Mark44 suggested you use [itex]cos^2(x)= 1- sin^2(x)[/itex] so that [itex]cos^2(x)- sin(x)= 0[/itex] becomes [itex]1- sin^2(x)- sin(x)= 0[/itex]. That is a quadratic equation in sin(x). If you let y= sin(x) you have [itex]y^2+ y- 1= 0[/itex]. Solve that that for y and then solve sin(x)= y for x.
OKAY so: f'' (x) = e^{sinx}(cos^{2}x - sinx) let cos2x - sinx = 0 1-sin^{2}(x) - sinx = 0 let y = sinx y^{2} + 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?
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 ??
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?
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?
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