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To determine the interval of increase/decrease and the min, we need to find the first derivative.
Since the first two factors are always positive the sign of f' depend on 3+x.
`f'(x)>0 => 3+x>0 => x>(-3)`
`f'(x)<0 => 3+x<0 => x<-3`
To find the point of inflections we set f'(x)=0 => x=0 or x=-3
From the graph we can see that x=-3 is the x-coord of the min. (We can also check that by using the 2nd derivative test)
Hence the abs min is `(-3,-27/e^3)`
The function increases over `(-3,oo)`
The function decreases over `(-oo,-3)`
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