Answer:
(sin x)^2*(sec x) is positive in QII
Step-by-step explanation:
(sin x)^2 is always 0 or positive. Here x lies in QII.
sec x is positive when the adjacent side is positive and negative when the adjacent side is negative. In QII the adjacent side is positive.
In summary, (sin x)^2*(sec x) is positive in QII
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<span>First we find the critical points by taking the first derivative and setting it equal to zero. So y'=27x^2-27=27(x^2-1)=27(x+1)(x-1)=0. The critical points are x=-1 and x=1. But we have a closed interval here and x=-1 is not in that interval. We must check the values of the endpoints and x=1, the critical point.
We substitute those back into the original and get
y(0)=9, y(1)=9-27+9=-9 and y(3)=9(3)^3-27*3+9=171.
When it asks for values, it is asking for y values.
The absolute minimum is -9 and the absolute maximum value is 171.</span>