SOLUTION: Can you please help me with these questions: Given f(x) = -1 if x<-2 2x +1 if -2 &#8804;x<1

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Question 1030187: Can you please help me with these questions:
Given f(x) = -1 if x<-2
2x +1 if -2 ≤x<1
x2 + x +1 if x ≥ 1
Is f(x) continuous:
i. As x tends to -2?
ii. As x tends to 1?
iii. Everywhere throughout its domain?
Question 2
Differentiate the following:
a. y=(x3 – 5x2)-3/2
b. f(x)=x¼ + x -3/2
---------------
X3 – 4x-5
c. [x4 – 3loge(x)][7ex+6x]
The demand function for the Deslynni wristwatch is given by
d(q) = 50 0 ≤ q < 20
------------------
0.02q2 + 1
Where q (measured in units of a thousand) is the quantity demanded per week and d(q) is the unit price in dollars.
i. Find d’(5) and interpret your results.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
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%28x%29+=+-1 if x+%3C+-2
The second piece of the piecewise function says f%28x%29+=+2x%2B1 if -2+%3C=+x%3C1
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%28-2%29+=+-1 but if we plug it into the second piece, we get f%28-2%29+=+-3. Since the outputs are NOT the same, the function is NOT continuous at x = -2

Work for the second piece

f%28x%29+=+2x%2B1

f%28-2%29+=+2%28-2%29%2B1 Replace every x with -2

f%28-2%29+=+-4%2B1

f%28-2%29+=+-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%28x%29+=+2x%2B1 if -2+%3C=+x%3C1
f%28x%29+=+x%5E2%2Bx%2B1 if x%3E=1

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

Second piece:
f%28x%29+=+2x%2B1
f%281%29+=+2%281%29%2B1
f%281%29+=+3

Third piece:
f%28x%29+=+x%5E2%2Bx%2B1
f%281%29+=+1%5E2%2B1%2B1
f%281%29+=+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.