Q. 31

Question

Sketch the graph of a function f described in Exercises 27–32, if possible. If it is not possible, explain why not.

$f h a s a r e m o v a b l e d i s c o n t i n u i t y a t x = - 2 a n d i s r i g h t c o n t i n u o u s a t x = - 2, a n d f (- 2) = 0 .$

Step-by-Step Solution

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Answer

Function not possible.

1Step 1. Given Information.

Given:

$f (- 2) = 0, f i s r i g h t c o n t i n u o u s a t x = - 2, a n d i t h a s a r e m o v a b l e d i s c o n t i n u i t y a t x = - 2 .$

2Step 2. Why function is not possible.

Now see that we are given that f has a removable discontinuity at $x = - 2$ this means the graph has a hole or a point at $x = - 2$ but just below in the second condition it is given that the absolute value of the function at $x = - 2$ is 0 which is a contradiction to it's own given condition.

Therefore, the function doesn't exist with these given conditions and so as the graph.

Q. 30

Q. 32

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Q. 29

Sketch the graph of a function f described in Exercises 27–32, if possible. If it is not possible, explain why not.f has a jump discon

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Q. 30

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Q. 32

Sketch the graph of a function f described in Exercises 27–32, if possible. If it is not possible, explain why not.f is continuous on [0, 2) but not on [0

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Q. 33

For each limit in Exercises 33–38, either use continuity to calculate the limit or explain why Theorem 1.16 does not apply.limx→-1 6.

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