What are the points of inflection of f(x)= sin^2x - cos^2x on x in [0, 2pi] ?

1 Answer
Sep 27, 2016

The points of inflection are :
(color(brown)(pi/4,0))Or(color(brown)((3pi)/4,0))

Explanation:

To find the inflection points we have to calculate the second derivative then set the equation to zero.

Knowing these properties:
color(blue)((sinu)'=u'cosu
color(red)((cosu)'=-u'sinu
color(green)(v^n=nv^(n-1))
color(orange)(sin2x=2sinxcosx

First let us find f''(x):

f'(x)=color(green)(2sinx(sinx)'-2(cosx)(cosx)') using 3rd property
f'(x)=2sinxcolor(blue)(cosx)-2cosx(color(red)(-sinx))
f'(x)=2sinxcosx+2cosxsinx
f'(x)=4sinxcosx
f'(x)=2color(orange)((2sinxcosx))
f'(x)=color(orange)(2sin2x)

Let us find f''(x):
f''(x)=(f'(x))'
f''(x)=(2sin2x)'
f''(x)=2color(blue)((sin2x))'
f"(x)=2*(color(blue)(2cos2x)) using first property above
f"(x)=4cos2x

Now set f"(x)=0
4cos2x=0
cos2x=0
In the given interval |0,2pi| then :
2x=(pi)/2rArrx=(pi)/4
f((pi)/4)=sin^2(pi/4)-cos^2(pi/4)=((sqrt2)/2)^2-(sqrt2/2)^2=0
The point is(color(brown)(pi/4,0))

Or
2x=(3pi)/2rArrx=(3pi)/4
f((3pi)/4)=sin^2((3pi)/4)-cos^2((3pi)/4)
f((3pi)/4)=(sin(pi-(pi)/4))^2-(cos(pi-(pi)/4))^2
f((3pi)/4)=(sin(pi/4))^2-(-cos(pi/4))^2
f((3pi)/4)=((sqrt2)/2)^2-((sqrt2)/2)^2=0
The second point is (color(brown)((3pi)/4,0))