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

- (x + 1)2 + 2; -o < x < -1

-+ 2; -1< x< 2

f(1) =

2

< 3 < 00

a

x= -1 and 2

b.

x= 2

c.

x = -1

d.

no points of discontinuity

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Option A.

Step-by-step explanation:

The given question is incomplete, here is the complete question.

Locate the points of discontinuity in the piece-wise function shown below.

f(x) = - (x + 1)² + 2     - ∞ < x < -2

        - x + 2            

Option A. x = -1 nd 2

            B. x = 2

            C. x = -1

            D. no points of discontinuity

Now to get the points of discontinuity we will check it at x = -1 first

For x = -1,

f(-1) = -(-1) + 2 = 3

If -(x+1)²+2 is true for x ≤ (-1)

f(-1) = -[(-1) + 1]² + 2

     = 0 + 2

Here we are getting two values of f(x) at x = (-1)

That means f(x) is not continuous at x = -1.

For x = 2    

For x ≥ 2

Then f(2) =

               = 1

If -x + 2 is true for x ≤ 2

f(2) = -2 + 2

     = 0

Again for x = 2 we are getting two values of f(2)

That clearly speaks that the function is not continuous at x = 2.

Therefore, option A is correct.