Please help identify which x-value is discontinuous and why ASAP, Thankyou!!

Question

Let f(x)= {​−2(x+3)^2 + 4, x ≤ -2

1/2x − 1,​ -2 < x < 2​ 

1/2√x-2, x > 2

For which x-value(s) is f discontinuous? Name the type of discontinuity and explain why the function is discontinuous at each x value using the definition of continuity. (Which of the 3-part definition is violated?)

Answer 1

-2(x+3)^2 +4 for x<-2

(1/2)x -1 for -2<x<2

(1/2(sqr(x)) -2 for x> 2

it's undefined at x=2, so immediately you can see that's a point of discontinuity (maybe skip down to the end of this post, to see the problem better)

but as to parts 1 and 2, the problem is a little more ambiguous depending on where you intended parentheses

parts 2 and 3 though are discontinuous regardless how you interpret part 3

-2(2+3)^2 +4 = -50+4=-46

(1/2)(-2) -1 = -2

it's discontinuous at x=-2

(1/2)sqr2 -2 = about 1.4/2 -2 = .7-2 =- 1.3

another discontinuity

1st part of f(x) is a downward opening parabola with vertex = (-3,4)

at -2 it's decreasing on the right side of the symmetrical parabola, at (-2,-46)

2nd part is a straight line segment from (-2,-2) to (2,0) upward sloping with slope=1

their is a huge "jump" a large discontinuity between the points (-2,0) and (-2,-2)

3rd part is another parabola, but horizontal and just the upper half of a rightward opening parabola

it comes closer to the line segment, but it's discontinuous too. (2,0) is the right end of the line segment and (2,-1.3) is the left end of the parabola

UNLESS, you meant 1/2sqr2 -2 as 1/2(sqr(2-2), 1/(2sqr2-2) or 1/2(sqr2-2) -2

and similarly 1/2(-2)-1 or (1/(2(-2)-1)

then 1/2sqr2-2 could = 1/0 which is undefined, or 1/.82 or 1/-2

and 1/2(-2)-1 = -5/4 or -1/5

while part 2's 1/2(-2) -1 = -5/4 or 1/(2(-2-1) = 1/-6

and 1/2(2)-1 = -3/4 or 1/(2(-2-1))= -1/6

IF you say parts 2 and 3 meet at (2,0) then you're interpreting part 2 as (1/2)x -1 so (1/2)(2)-1 =0

but part 3 would be (1/2)sqr(2-2) = 1/2)sqr0 = 0

which appears continuous, EXCEPT the graph is undefined at x=2

UNLESS you left out an equal sign in part 2 or 3, but as is, there is no defined value of f(x) when x=2

it's "very small" discontinuity, a very small "hole" in the graph, but it's still discontinuous. a tiny "hole" of infinitesmal diameter.