## anonymous one year ago use the definition of continuity to determine whether f is continuous at a.

1. anonymous

2. perl

I can't read that because i don't have microsoft word. one moment

3. anonymous

ok

4. perl

5. anonymous

ok thank you so much for helping me

6. perl

7. anonymous

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

8. perl

for continuity we need to check the points where the function changes, the rest of the function is continuous in its respective parts.

9. anonymous

ok how do we do that

10. perl

Definition of continuity: Left hand limit = right hand limit = value at the point

11. perl

We can write this in symbols.

12. anonymous

you write it with a little apostrophe thing right

13. perl

Definition of continuity: $$\Large \lim_{x \to a} f(x) = f(a)$$But this implies $$\Large \lim_{x \to a^{+}} f(x) = \Large \lim_{x \to a^{-}} f(x) = f(a)$$

14. perl

So we need to check the left limit and right limits and make sure it equals to the value at that point.

15. perl

$$\rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} }$$

16. anonymous

for all the x's i replace them with a 1?

17. perl

$\rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} }$ $\Large \lim_{x \to 1^{+}} f(x) = \\~\\ \Large \lim_{x \to 1^{-}} f(x) = \\~\\ \Large f(1) =$

18. perl

yes thats correct. and we need to check that all three of those are equal . thats the continuity condition

19. anonymous

so the answer is 1 for all of them.

20. perl

$\rm \Large{f(x)= \begin{cases} 2-x ~~~if~ x<1\\ 1 ~~~~~~~~ ~\rm ~if~ x=1\\ x^2 \rm ~~~~~~~~if~ x>1 \end{cases} }$ $\Large \lim_{x \to 1^{-}} f(x) =\lim_{x \to 1^{-}} 2-x = 2-1 = 1 \\~\\ \Large \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} x^2 = 1^2 = 1 \\~\\ \Large f(1) = 1$

21. anonymous

Thank You So Much,can you help me with one more similar to this one.

22. perl

ok

23. anonymous

determine for what numbers, if any, the given function is discontinuous. f(x)={x+7 if x is less than or equal to 0 {7 if 0 is <x less than or equal to 3 {x^2-1 if x>3

24. anonymous

ok whats the difference between the first set and the second set. (not the questions but what you did)

25. perl

there are two potential points of discontinuity, where the function abruptly changes. it changes at x =0 and x = 3. so we have to test those two parts independently

26. anonymous

ok so how do we do that

27. perl

We can look at the graph that might make it more clear https://www.desmos.com/calculator/pcjrrf5xdh

28. perl

edit* $\rm \Large{f(x)= \begin{cases} x+7 ~~~if~ x\leq0\\ 7 ~~~~~~~~ ~\rm ~if~ 0 < x\leq 3\\ x^2-1 \rm ~~if~ x>3 \end{cases} }$ $\Large \lim_{x \to 0^{-}} f(x) =\lim_{x \to 0^{-}} x+7 = 0+7 = 7 \\~\\ \Large \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} 7 = 7 \\~\\ \Large f(0) =0+7 = 7$ $\Large \lim_{x \to 3^{-}} f(x) =\lim_{x \to 3^{-}}7 = 7 \\~\\ \Large \lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} x^2-1 = 3^2-1 = 8 \\~\\ \Large f(3) = 7$

29. perl

from the graph and from the continuity definition we see that there is a point of discontinuity at x = 3. x=0 is fine

30. anonymous

so all the question was asking for was two points

31. perl

Before we do any work the function is potentially discontinuous at the two points x=0 and x = 3 because the function changes suddenly at those points. But after analyzing the function graphically, or using the limit above, we see that it is actually only discontinuous at one point. x= 3