I think at this point the "analytical" thing to do would be to set each side equivalent to 0 and solve. So that:
1+sin(x)=0
sin(x)=-1
x=3\pi/2
2sin(x)cos(x)=0
sin(x)=0
x=\pi and 2\pi
2sin(x)cos(x)=0
cos(x)=0
x=3\pi/2
At this point, I think it becomes a plug-and-chug with 3\pi/2...
@Mark44:
I agree that inspection of the graph is easier for this problem, but my professor likes for us to show critical numbers.
@xaos:
Once you pointed out the evaluation at 1 (rather than -1), a typo seems to be the most logical conclusion.Respectfully to all,
Aaron
"Calculus" (Larson, et al) 9th ed: p. 169 #29: No match to answer key.
Homework Statement
Locate the abs extrema on the interval of the function:
y=t-|t-3| for interval [-1,5]
Homework Equations
|x|=\sqrt{x^{2}}
The Attempt at a Solution
I thought this would essentially be a...