What are the points of inflection of #f(x)= 3sin2x - 2xcosx# on #x in [0, 2pi]
#?

Question

What are the points of inflection of #f(x)= 3sin2x - 2xcosx# on #x in [0, 2pi]
  #?

Emma Aldridge · Accepted Answer

x #approx {0, 1.381, 2.931, 4.831}#

Find the first and second derivatives first.

#f(x) = 3sin(2x)-2xcos(x)# #f'(x) = 2xsin(x)+6cos(2x)-2cos(x)# #f''(x) = 2xcos(x) + 4sin(x) - 12sin(2x)#

Next go ahead and set #f''(x)# equal to zero

#2xcos(x) + 4sin(x) - 24sin(x)cos(x) = 0# (However I am unable to solve this equation for y = 0, expect for x = 0, if it is solvable at all)

Looking at the graphs of #f(x)#, #f'(x)#, and #f''(x)# we can make estimations of the zeros on the interval #[0, 2pi]#

#f(x)# graph{3sin(2x)-2xcos(x) [0, 6.2831853071795864769, -5, 5]} #f'(x)# graph{ 2xsin(x)+6cos(2x)-2cos(x) [0, 6.2831853071795864769, -5, 5]} #f''(x)# graph{2xcos(x) + 4sin(x) - 12sin(2x) [0,6.2831853071795864769, -5, 5]}

For x #approx {0, 1.381, 2.931, 4.831}# we see that #f''(x)# is zero and that #f'(x)# has maximums and minimums as well as #f(x)# appears to be changing concavity.

Ezra Adams · Answer

undefined To find the points of inflection, we first need to find the second derivative of the function and then solve for the values of $ x $ where the second derivative changes sign within the interval $ [0, 2\pi] $.

First derivative: $ f'(x) = 6\sin(x)\cos(x) - 2\cos(x) + 2x\sin(x) $

Second derivative: $ f''(x) = 6\cos^2(x) - 6\sin^2(x) - 2\sin(x) + 2\sin(x) + 2x\cos(x) + 2\sin(x) $

Simplify the second derivative: $ f''(x) = 6(\cos^2(x) - \sin^2(x)) + 2x\cos(x) + 2\sin(x) $

Using the trigonometric identity $ \cos(2x) = \cos^2(x) - \sin^2(x) $, we rewrite the second derivative: $ f''(x) = 6\cos(2x) + 2x\cos(x) + 2\sin(x) $

Now, find the points of inflection by solving $ f''(x) = 0 $ and $ f''(x) $ changes sign within $ [0, 2\pi] $.

This involves finding the solutions of $ 6\cos(2x) + 2x\cos(x) + 2\sin(x) = 0 $ within the interval $ [0, 2\pi] $.

There are multiple methods to solve this equation, such as numerical methods or graphical analysis. Once you find the solutions within the interval $ [0, 2\pi] $, those are the points of inflection for the function $ f(x) $.

What are the points of inflection of #f(x)= 3sin2x - 2xcosx# on #x in [0, 2pi] #?