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Locate the points of discontinuity in the piecewise function shown below.

f(x) = {-(x+1)² + 2; -[infinity] < x < -1
{-x+2; -1 ≤ x < -1
{√x-1 ; 2 ≤ x < [infinity]

a. x = -1 and 2
b. x = 2
c. x = -1
d. no points of discountinuity

User Dilfish
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Final answer:

The points of discontinuity in the piecewise function are at x = -1 and x = 2 because the function changes definition at these points, with the second part not valid for any x and the third part starting from x = 2.

Step-by-step explanation:

The question asks us to find the points of discontinuity of the given piecewise function. A point of discontinuity is where a function is not continuous. In our case, we're looking at a function that is defined differently for different intervals of the domain. The first part of the function, -(x+1)² + 2, is valid for x values strictly less than -1. The second part of the function, -x+2, is only defined for -1. The third part, √x-1, starts from x values of 2 and goes onward.

To find discontinuities, we check where the function definition changes. At x = -1, the function switches from the first to the second part. However, the second part is not valid for any x as there's a typo (-1 < x < -1 is an impossible interval). The third part starts from x = 2. So, we identify that the points of discontinuity are x = -1 and x = 2. At x = -1, the definition of the function changes but does not provide any valid output in the second interval, which makes it discontinuous. At x = 2, the function jumps from being undefined to the definition provided by the third part of the piecewise function, thereby creating another discontinuity.

User DeutschZuid
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