Consider the function f (x) = StartLayout Enlarged left-brace first row negative 4 x squared, x less-than-or-equal-to negative 1 second row negative 3 Superscript negative x Baseline minus 1, x greater-than negative 1 EndLayout. Which statement explains the continuity of the function at x = –1?

Because Limit as x approaches negative 1 f(x) = 4 and f(–1) = 4, it follows that Limit as x approaches negative 1 f(x) = f(–1). Therefore, the function is continuous at x = –1.
Because Limit as x approaches negative 1 f(x) = –4 and f(–1) = –4, it follows that Limit as x approaches negative 1 f(x)= f(–1). Therefore, the function is continuous at x = –1.
Because Limit as x approaches negative 1 f(x) = 4 and f(–1) = –4, it follows that Limit as x approaches negative 1 f(x) ≠ f(–1). Therefore, the function is not continuous at x = –1.
Because Limit as x approaches negative 1 f(x) = –4 and f(–1) = 4, it follows that Limit as x approaches negative 1 f(x) ≠ f(–1). Therefore, the function is not continuous at x = –1.

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step-by-step explanation:

faster, im not sure why but i know it is faster

then say something, tell them they won’t respect you if you don’t even have respect for your self sweetheart you need to lay it on them, all of it : )

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.