2. Ifj(x) = x2 + 1, find j(a + 1).
1 answer:
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Go to the left-most point on the graph. That point is (-3,-2). The x coordinate of this point is x = -3.
So the smallest x can be is x = -3. The variable x can't be any smaller. It can be larger than -3.
Since x can equal -3 or be larger, this makes the domain to be
(x is greater than or equal to -3)
In interval notation, you would write
The square bracket tells you to include the endpoint -3 while the curved parenthesis tells you to exclude infinity as an endpoint. Plus or minus infinity is always associated with a curved parenthesis.
In short, the answer is choice D
So the smallest x can be is x = -3. The variable x can't be any smaller. It can be larger than -3.
Since x can equal -3 or be larger, this makes the domain to be
In interval notation, you would write
The square bracket tells you to include the endpoint -3 while the curved parenthesis tells you to exclude infinity as an endpoint. Plus or minus infinity is always associated with a curved parenthesis.
In short, the answer is choice D