# Second derivative/Critical Numbers of a trigonometric function

I was attempting to find the inflection points and concavity of this function but my answers have been repeatedly wrong. I believe it may have something to do with how I did my second derivative. I was wondering if anyone could help. Thank you.

f(x) = 3cos^2(x) - 6 sinx, 0<= x <= 2{/pi}

I already found out that the first derivative would be:

f'(x) = -6sincosx - 6cosx

I'm not 100% sure of the second derivative and its critical numbers though.

Edit: My original second derivative was -6(cos^2(x) - sin^2(x) - cos(x)) though, if this is right, then I'm having trouble finding the correct critical numbers.

• How can you get the first derivative but not be able to get the second derivative? You can just use the same methods. – Suzu Hirose Nov 13 '14 at 4:55
• The first derivative is $0$ at $\pi/2$, $3\pi/2$. – André Nicolas Nov 13 '14 at 5:16
• Right. I have that. What I'm missing is when the second derivative is 0. – user3495234 Nov 13 '14 at 5:16

If $$f(x)=3 \cos ^2(x)-6 \sin (x)$$ $$f'(x)=-6 \cos (x)-6 \sin (x) \cos (x)=-6 \big(\sin (x)+1\big) \cos (x)$$ $$f''(x)=6 \sin ^2(x)+6 \sin (x)-6 \cos ^2(x)=6\big(2\sin^2(x)+\sin(x)-1\big)$$ Notice that you made a small mistake for the second derivative.
As commented by André Nicolas, the first derivative cancels at $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$.
If you consider now equation $2z^2+z-1=0$, its roots are $z=-1$ and $z=\frac{1}{2}$ which means that the second derivative cancels when, in the range, $\sin(x)=-1$ and $\sin(x)=\frac{1}{2}$.