Respuesta :

[tex]f(x)=-\dfrac{x^2}2+x-\cos^2x[/tex]

[tex]\implies~f'(x)=-x+1-2\cos x(-\sin x)=-x+\sin2x+1[/tex]

[tex]\implies~f''(x)=-1+2\cos2x[/tex]

Inflection points always occur at points where the second derivative is equal to zero, but not every such point will be an actual inflection point.

[tex]-1+2\cos2x=0\iff\dfrac12=\cos2x[/tex]
[tex]\implies 2x=\pm\dfrac\pi3+2n\pi[/tex]
[tex]\implies x=\pm\dfrac\pi6+n\pi[/tex]

where [tex]n[/tex] is an integer. There are infinitely many inflection points for this function.