Question 1030187
I'll answer the first question. Please only post one question per post. Thank you.


Part i)


The first piece of the piecewise function says {{{f(x) = -1}}} if {{{x < -2}}}
The second piece of the piecewise function says {{{f(x) = 2x+1}}} if {{{-2 <= x<1}}}
we're focusing on the first and second pieces because x = -2 is at the junction between the two pieces.


Plug in x = -2 into each piece. If we plug x = -2 into the first piece, we get {{{f(-2) = -1}}} but if we plug it into the second piece, we get {{{f(-2) = -3}}}. Since the outputs are NOT the same, the function is NOT continuous at x = -2


Work for the second piece


{{{f(x) = 2x+1}}}


{{{f(-2) = 2(-2)+1}}} Replace every x with -2


{{{f(-2) = -4+1}}}


{{{f(-2) = -3}}}


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Part ii)


now we're focusing on the second and third pieces because x = 1 is at the junction between the two pieces.


The second and third pieces of the piecewise function are...


{{{f(x) = 2x+1}}} if {{{-2 <= x<1}}}
{{{f(x) = x^2+x+1}}} if {{{x>=1}}}


Let's plug in x = 1 to see what happens


Second piece:
{{{f(x) = 2x+1}}}
{{{f(1) = 2(1)+1}}}
{{{f(1) = 3}}}


Third piece:
{{{f(x) = x^2+x+1}}}
{{{f(1) = 1^2+1+1}}}
{{{f(1) = 3}}}


Since the outputs are the same, this function is continuous at x = 1.


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Part iii)


Since back in part i) I've shown that the function is NOT continuous at x = -2, this means that the function is NOT continuous throughout its domain.