f(x) = 2sinx - cos^2x
rArr f'(x) = 2cosx - 2*cosx*sinx (by the chain rule).
= 2cosx - sin(2x) (by the double-angle formula for sin).
rArr f''(x) = -2sinx - 2cos(2x)
= -2(sinx - cos(2x))
One of the conditions for a point of inflexion at x=a is that f''(a) =0 ... however, f''(x) must also change sign.
First we solve the equation f''(x) = 0
= -2(sinx - cos(2x)) = 0
rArr sinx - cos(2x) = 0
Expanding cos(2x) with one of the double-angle formulas for cos:
sinx - (1 - 2sin^2x) = 0
:. 2sin^2x + sinx - 1 = 0
Now we let u = sinx
2u^2 + u - 1 =0
rArr u = -1 or u = 1/2 (by factorising the quadratic).
rArr x in {pi/6, (5pi)/6, (3pi)/2} (taking the inverse sine of the previous u-values in the given domain).
Now, to ensure that f''(x) changes sign across these x-values, we can just evaluate the following:
f''(0) = 2
f''(pi/2) = - 4
f''(pi) =2
f''(2pi) = 2
This shows that f''(x) changes sign across x=pi/6 and x=(5pi)/6 but not x=(3pi)/2, so the there is no point of inflexion at x=(3pi)/2
So, the co-ordinates of the points of inflexion for f(x) are:
(pi/6, 1/4) and ((5pi)/6, 1/4)
Hope this helps!
