critical and inflection pts

• Nov 8th 2007, 06:45 AM
coe236
critical and inflection pts
find the critical points of f(x) and use the second derivative test (if possible) to determine whether each corresponds to a max or min.

$f(x)=sin(x)^2+cos(x)$ ,[0,pi]

i got f'(x) and f''(x) but im just having a little trouble setting them equal to zero and solving for the critical and inflection points. Could someone help me out on that please?

$f'(x)=2*sin(x)*cos(x)-sin(x)=0$ x=? -->critical pts.
$f''(x)=4*cos(x)^2-cos(x)-2=0$ x=? -->inflection pts
• Nov 8th 2007, 06:51 AM
coe236
i got the critical pts actually, x=pi/3 and x=0
and for the inlfection pts its cosx=2 and cosx=3/4 ..is that possible?
• Nov 8th 2007, 06:52 AM
PaulRS
For the zeros in the first derivative factorize.

$2\cdot{\sin(x)\cdot{\cos(x)}}-\sin(x)=0$

Then $\sin(x)\cdot{\left(2\cdot{\cos(x)}-1\right)}=0$

From there you get $\sin(x)=0$ and $2\cdot{\cos(x)}-1=0$

For the second derivative

Try this change of variable $z=\cos(x)$ what do you see? :D
• Nov 8th 2007, 06:58 AM
coe236
i understand now, thankyou !