Points of discontinuity?

Points of discontinuity are where the function is undefined.

#6)#

#y = 6^(1/(x+3))+2# is only undefined when #x -> -3^+# or #x -> -3^(-)#, because #1/0# from the left or right goes to #pmoo#. For #x = -3# you would have a vertical asymptote.

The range can be defined.

• Below #x = -3#, the function is increasing from right to left towards #y = 3# because #6^(1//-"large") = 1# and #1 + 2 = 3#. Approaching #x = -3#, #6^(1//-0.0000001) = 6^(-oo) = 0#, so from the left, #y -> 2#.

• Above #x = -3#, #6^(1//0.0000001) = 6^(oo) = oo#, so the function starts high. As #x->oo#, #y -> 6^(1//oo) = 1#, and #1+2 = 3#, so the function flattens out to give #y = 3# as #x->oo#.

Therefore, the graph can be sketched as:

graph{6^(1/(3+x))+2 [-9.15, 8, 0.075, 7.725]}

#7)#

#y = {(5, x<=-2),(x^2-2x-3, -2 < x < 3 ),(3x-7, x >= 3) :}#

This is a piece-wise function. You just have to sketch each portion in that domain.

• #y(-2) = 5# from the left, and #(-2)^2 - 2(-2) - 3 = 5# from the right, so that is a continuous connection.
• #y(3) = (3)^2 - 2(3) - 3 = 0# from the left, and #3(3) - 7 = 2# from the right, so there is a jump discontinuity at #x = 3#. Only the #3x - 7# point exists due to the fully open domain of the middle function.
• All three functions are continuous in their own domains.

Therefore, only one discontinuity exists. I'll let you sketch this graph... Socratic has a hard time graphing piecewise functions like this.