Piece function: x^2-5: x> 3 -2x+8: x less than 3 5: x=3 a)lim 3+ > i got 1 b) lim 3- > i got 2 c) lim 3 > i got 5 d) f(3) > i put 5 e) therefore is f(x) continuous at x = 3 explain. my no because the limits from both sides are different

First, make the table so you can sketch the graph:

    x   |    y

    5  |   20        

    4  |    9              x² - 5   ; x > 3

    3  |    4  (open dot)                         ⇒ from the right

   3  |    5                  5     ; x = 3         ⇒ at x = 3

    3  |    2  (open dot)                          ⇒ from the left

    2  |    4              -2x + 8  ; x < 3          

    1   |    6                                          

Next, look at the graph (or table) to find the limits:

lim 3⁺ = 4  as x approaches 3 from the right, y approaches 4

lim 3⁻ = 2  as x approaches 3 from the left, y approaches 2

lim 3 = DNE lim 3⁺ ≠ lim 3⁻ so the limit does not exist

f(3) = 5  when x = 3, y = 5

f(x) is NOT continuous at x = 3 because lim 3 ≠ f(3)

three-fourths of an hour

36÷48=0.75

0.75=

the square root of what number? if you add that, then someone will probably answer the question.