3 for r 1 2 15 pts Graph the function f r 3 for 1 r 1 13 a F

3 for r -1 2. (15 pts.) Graph the function f (r) 3 for -1 r 1 13 a. Find the domain of DE (-oo, oa) f (r) b. Find f (-2) c. Find f (3) not, what type of discontinuity exists? d. Is the function continuos at z l? If e. Is the function continuos at z l? If not, what type of discontinuity exists? f, Is the function continuos at a 5? If not, what type of discontinuity exists?

Solution

3 - x ; x < -1
x^2 + 3 ; -1 <= x < 1
1/(5-x) ; 1 <= x < 5

Notice the intervals above...
The transition values x = -1 and x = 1 are included

And also, 1/(5-x) is discontinuous at x = 5
But 5 aint included in the domain 1 <= x < 5 after all

So, the domain is :
[-1 , 5)

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f(-2) :
This satisfies x < -1 for
which f(x) = 3 - x
3 - (-2)
3 + 2
5 ---> ANSWER

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f(3) :
3 satisfies 1 <= x < 5
for which f(x) = 1/(5 - x)
1/(5 - 3)
1/2
0.5 ----> ANSWER

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d) Plug in x = -1 into the first two functions...
3 - (-1) = 3 + 1 ---> 4
x^2 + 3 --> (-1)^2 + 3 ---> 1 + 3 ---> 4
They are equal
So, the function is continuous at x = -1

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e) Plug in x = 1 into the second and third function :
x^2 + 3 ---> 1^2 + 3 ---> 1 + 3 ---> 4
1/(5-x) --> 1/(5-1) ---> 1/4
they are unequal, so discontinuous at x = 1
Thetype of discontinuity is a JUMP discontinuity

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f) x = 5 :
Notice the domain is [-1 , 5)
5 aint included in the domain
So, when we test for continuity at x = 5, we only look at the left hand limit

1/(5 - x)
As x approaches 5 from the left, 1/(5-x) approaches +infinity

So, discontinuous at x = 5

INFINITE DISCONTINUITY is the type