The intervals refer to the x values where the function (graph) is increasing which means the y values are getting larger from left to right. Parantheses are used not brackets (parantheses means it doesn't include the value and brackets mean it does) so here the intervals would be
( negative infinity , -3)U( -3, 0 )
Answer:
Step-by-step explanation: I think it is c
Answer:
The absolute maximum is
and the absolute minimum value is 
Step-by-step explanation:
Differentiate of
both sides w.r.t.
,


Now take 



![\Rightarrow 1-2\sin ^2t =\sin t \quad \quad [\because \cos 2t = 1-2\sin ^2t]](https://tex.z-dn.net/?f=%5CRightarrow%201-2%5Csin%20%5E2t%20%3D%5Csin%20t%20%20%5Cquad%20%5Cquad%20%20%5B%5Cbecause%20%5Ccos%202t%20%3D%201-2%5Csin%20%5E2t%5D)






In the interval
, the answer to this problem is 
Now find the second derivative of
w.r.t.
,

![\Rightarrow \left[f''(t)\right]_{t=\frac {\pi}6}=-2\times \frac {\sqrt 3}2-4\times \frac{\sqrt 3}2=-3\sqrt 3](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cleft%5Bf%27%27%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D-2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2-4%5Ctimes%20%5Cfrac%7B%5Csqrt%203%7D2%3D-3%5Csqrt%203)
Thus,
is maximum at
and minimum at 
![\left[f(t)\right]_{t=\frac {\pi}6}=2\times \frac {\sqrt 3}2+\frac{\sqrt 3}2=\frac{3\sqrt 3}2\;\text{and}\;\left[f(t)\right]_{t=\frac{\pi}2}= 2\times 0+0=0](https://tex.z-dn.net/?f=%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%20%7B%5Cpi%7D6%7D%3D2%5Ctimes%20%5Cfrac%20%7B%5Csqrt%203%7D2%2B%5Cfrac%7B%5Csqrt%203%7D2%3D%5Cfrac%7B3%5Csqrt%203%7D2%5C%3B%5Ctext%7Band%7D%5C%3B%5Cleft%5Bf%28t%29%5Cright%5D_%7Bt%3D%5Cfrac%7B%5Cpi%7D2%7D%3D%202%5Ctimes%200%2B0%3D0)
Hence, the absolute maximum is
and the absolute minimum value is
.