## amy0799 one year ago consider the following. f(x) = (x)/(x^2-2) Find the x-values at which f is not continuous, smaller value = larger value = Is the discontinuity at the smaller value removable? Is the discontinuity at the larger value removable?

1. anonymous

look for a value of x that makes the denominator zero, it won't be continuous there

2. Michele_Laino

hint: for continuity, we can not divide by zero

3. amy0799

i'm confused on what to do...

4. Michele_Laino

we have tosolve this equation, first: $\Large {x^2} - 2 = 0$

5. amy0799

x=sqrt(2)

6. Michele_Laino

more precisely, we have two solutions: $\Large x = \pm \sqrt 2$

7. Michele_Laino

am I right?

8. amy0799

yea. so the smaller value is $-\sqrt{2}$ and the larger value is $\sqrt{2}$?

9. Michele_Laino

Correct! we have to exclude both values, since we can not divide by zero, so our function is not continuous at these points: $\Large \begin{gathered} x = \sqrt 2 \hfill \\ x = - \sqrt 2 \hfill \\ \end{gathered}$

10. amy0799

right. so how do you find out if it's removable?

11. Michele_Laino

I think that both discontinuities are not removable

12. amy0799

can you explain why?

13. Michele_Laino

yes! I explain with an example

14. Michele_Laino

for example, the subsequent function: $\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}}$ is not continuous at x=1, right?

15. amy0799

yes

16. Michele_Laino

nevertheless x=1 is a removable discontinuity, since we can write this: $\Large f\left( x \right) = \frac{{{x^2} - 1}}{{x - 1}} = \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = x + 1$

17. amy0799

ooh ok. can you help me with another problem?

18. Michele_Laino

yes!

19. amy0799

consider the following. f(x) = {-9x

20. Michele_Laino

ok!

21. amy0799

f(x) = -9 if x less than or equal to 5 x^2-2x+2 if x greater than 5

22. amy0799

find the x-value at which f is not continuous and is the discontinuity removable

23. Michele_Laino

at x= 5, we have f(x)= 9 using your first part for f(x) and we have: ${x^2} - 2x + 2 = 25 - 2 \cdot 5 + 2 = 17 \ne 9$ so the discontinuity is at x=5

24. amy0799

is it not removable?

25. Michele_Laino

I think that it is not removable

26. Michele_Laino

because it is a jump-type discontinuity

27. amy0799

ok thanks!. I have a few more I need help me, is that ok?

28. Michele_Laino

ok!

29. amy0799

g(x) = 5sin(x)/x if x less than 0 a-4x if x greater than or less than 0 Find the constant a such that the function is continuous on the entire real line.

30. Michele_Laino

we can write this: $\Large \mathop {\lim }\limits_{x \to 0} \frac{{5\sin x}}{x} = 5$

31. Michele_Laino

whereas at x=0, we have: $\Large a - 4x = a - 4 \cdot 0 = a$

32. Michele_Laino

now, for continuity, we have to request taht both values are equal each to other. So what do you get?

33. amy0799

Would the constant a be 5?

34. Michele_Laino

that*

35. Michele_Laino

correct! a=5

36. amy0799

g(x) = (x^2-a^2)/(x-a) if x is not equal to a 11 if x=a Find the constant a such that the function is continuous on the entire real line.

37. Michele_Laino

we can write this: $\Large \frac{{{x^2} - {a^2}}}{{x - a}} = \frac{{\left( {x - a} \right)\left( {x + a} \right)}}{{x - a}} = x + a$

38. Michele_Laino

and its value at x=a, is: $\Large 2a$ Now, for continuity we have to request, as before, that those two values are equal each to other, so we have to solve this equation: $\Large 2a = 11$

39. amy0799

I don't understand how you got 2a

40. Michele_Laino

after a simplification, I got this: $\Large g\left( x \right) = x + a$ so, $\Large g\left( a \right) = a + a = 2a$

41. amy0799

oh ok. so the constant a is 11/2?

42. Michele_Laino

yes! correct!

43. amy0799

Describe the interval(s) on which the function below is continuous. $f(x)=x \sqrt{x+1}$

44. Michele_Laino

your function is continuous inside all its domain, and its domain is the subset of all real numbers, such that the radicand is positive or equal to zero, namely: $\Large x + 1 \geqslant 0$

45. amy0799

[0,∞) is that how you would write it?

46. Michele_Laino

we have: $\Large x + 1 \geqslant 0 \Rightarrow x \geqslant - 1$ so, the requested interval is: $\Large [ - 1,\infty )$ am I right?

47. amy0799

yes. Thank you so much for the help and sorry if I took up your time.

48. Michele_Laino

no worries! :) :)