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Question 987496: For the function f(x) = 6-(1/x+3), answer the following questions:
A) Domain and range
B) 1 to 1 or not
C) Continuous or not (if continuous tell where)
D) Increasing, decreasing, both (if both tell where), or neither
E) Bounded above, bounded below, bounded, or unbounded (if unbounded, tell where)
F) Roots
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I am going to write this as 6- [1/(x+3)]
That is very different from 6-(1/x) +3
The domain is all reals except x= -3
Range: look at x approaching -3 from the + side and from the - side.
As x approaches from the + side, the fraction gets large positive 1/(-2.999+3)=1/.001=1000, and that makes the function large negative, since it is being subtracted.
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And by the same approach from the negative side, it can be infinite negative as well, which subtracted, makes the range infinitely positive.
It is DISCONTINUOUS at x=-3
It increases from minus infinity to 6 and decreases from positive infinity to 6 as x gets large negative. The bounds are asymptote is 6. As x gets large positive, the function goes from negative infinity to 6.
It is 1 to 1, for while it doesn't look like it passes either horizontal or vertical line test, the function does change slightly.
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Solving for roots, -1/(x+3)=-6
6x+18=1
6x=-17
x= -(17/6)
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