Answer:
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Step-by-step explanation:
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Please only post one problem at a time. I'll answer problem 1 to get you started. If you still need help with the rest, then post them as separate posts and I'll have a look at them.
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Notice how x = -1 is the junction point, where one piece transitions to the other
Plug x = -1 into each piece of the piecewise function
So for the first piece we have
b - 2x
b - 2(-1)
b + 2
and similarly for the second piece
-6/(x-b)
-6/(-1-b)
Now set the two expressions equal to each other and solve for b
-6/(-1-b) = b+2
-6 = (b+2)(-1-b)
(b+2)(-1-b) = -6
-b-b^2-2-2b = -6
-b-b^2-2-2b+6 = 0
-b^2-3b+4 = 0
b^2+3b-4 = 0
(b-1)(b+4) = 0
b-1 = 0 or b+4 = 0
b = 1 or b = -4
The value of b with greater absolute value is b = -4 as this value is further from zero (compared to b = 1)
So -4 goes in the box
This is what the graph of f looks like (see attached) when b = -4
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Notice how x = -1 is the junction point, where one piece transitions to the other
Plug x = -1 into each piece of the piecewise function
So for the first piece we have
b - 2x
b - 2(-1)
b + 2
and similarly for the second piece
-6/(x-b)
-6/(-1-b)
Now set the two expressions equal to each other and solve for b
-6/(-1-b) = b+2
-6 = (b+2)(-1-b)
(b+2)(-1-b) = -6
-b-b^2-2-2b = -6
-b-b^2-2-2b+6 = 0
-b^2-3b+4 = 0
b^2+3b-4 = 0
(b-1)(b+4) = 0
b-1 = 0 or b+4 = 0
b = 1 or b = -4
The value of b with greater absolute value is b = -4 as this value is further from zero (compared to b = 1)
So -4 goes in the box
This is what the graph of f looks like (see attached) when b = -4