Question

Problem 3 in attached image

Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

Possible discontinuities of g(x) are at x=0 and x=1.

We find left and right hand limits at x=0 and x=1.

Step 2

At x=0, left hand limit=0+1=1

And right hand limit = e^0=1

So at x=0, g(x) is continuous.

At x=1 left hand limit=e^1=e

Right hand limit=2-1=1

So, at x=1 the graph is discontinuous.

Answer(a): x=1

Step 3

g(1)= left hand of g(x) limit at x=1

Answer(b)...

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