## anonymous one year ago Need Help! Find where the function f(x)=x+sqrt(1-x) is increasing and decreasing.

1. anonymous

|dw:1439260678254:dw|

2. anonymous

step one find the derivative

3. anonymous

|dw:1439260757824:dw|

4. anonymous

let me know when you get $f'(x)=1-\frac{1}{2\sqrt{1-x}}$

5. anonymous

I'm just having trouble trying to solve for x

6. anonymous

ok lets back up a second

7. anonymous

exponential notation is great for computing certain derivatives, but it totally sucks for doing actual numeric computation

8. anonymous

the square root function is a very very very common function so you should commit its derivative to memory and not screw around with rational exponents $\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}$ like memorizing $$6\times 9=54$$

9. anonymous

and via the chain rule $\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}$

10. anonymous

I didnt learn it that way, I was taught to set it up with the 1/2 in the front and -1/2 as the exponent value. But if doing it the way above helps me solve the problem, ill stick with it

11. anonymous

math teachers love to put the square root function on a test, quiz etc so while your colleages are converting to exponents, then using the power rule etc, you should go right to $f'(x)=1-\frac{1}{\sqrt{1-x}}$

12. anonymous

yeah the power rule works for sure, but the derivative never changes

13. anonymous

so once you know it, you know it like $$8\times 7=56$$ it is always the same and yes, you need to do any computation at all with a derivative, you have to get rid of the rational exponent and write what it actually is

14. anonymous

|dw:1439261266064:dw|So would the function look like this

15. anonymous

now you can solve $1-\frac{1}{2\sqrt{1-x}}=0$ to find the critical points etc you cannot do it with $1-\frac{1}{2}(1-x)^{-\frac{1}{2}}$that form i useless

16. anonymous

yeah what you wrote now we can do it

17. anonymous

Do we multiple the 1 by the denominator of the other fraction?

18. anonymous

you want to set it equal to zero and solve probably a good first step

19. anonymous

oh kk

20. anonymous

notice that the deriviative has domain $$x<1$$ which is not surprising because the original function has domain $$x\leq 1$$

21. anonymous

also notice that the derivative is strictly decreasing

22. anonymous

So does x equal 3/4 ?

23. anonymous

which means all that is left to do is solve $1-\frac{1}{2\sqrt{x-1}}=0$

24. anonymous

yes it does

25. anonymous

Okay, so then I would make a number line and test points on the left and the right and see where it is postive and negative which should give me where the function is inc or dec?

26. anonymous

so $$f'(x)<0$$ i.e is negative if $$x\geq \frac{3}{4}$$ and $$f'(x)>0$$ i.e positive of $$x\leq\frac{3}{4}$$

27. anonymous

yeah you can do that if it is not obvious

28. anonymous

then translate to increasing decreasing for $$f$$ and you are done

29. anonymous

quick quiz $\frac{d}{dx}[\sqrt{1-x^2}]=?$

30. anonymous

so (-infinity, 3/4) is incr and decreasing from (3/4, infinity)

31. anonymous

oh no hold the phone

32. anonymous

increasing on $$(-\infty, \frac{3}{4})$$ is right

33. anonymous

1/2*sqrt(1-x^2) *-2

34. anonymous

1/(2*sqrt(1-x^2))*-2

35. anonymous

close the numerator should be $$-2x$$

36. anonymous

and of course the two's cancel

37. anonymous

Oh yea, was in a rush

38. anonymous

ok lets get back to your question

39. anonymous

$f(x)=x+\sqrt{1-x}$right?

40. anonymous

is it undefined from 3/4 to infinity

41. anonymous

don't write that is increasing on $$(\frac{3}{4},\infty)$$ or your teacher will think you are daft

42. anonymous

the domain is $(-\infty, 1]$ right?

43. anonymous

right so we just dont include the 3/4 to infinity part, right?!

44. anonymous

so it is only increasing on $(\frac{3}{4},1)$

45. anonymous

yeah you include it , but dont to to infinity

46. anonymous

three fourth is less than one, so there is an interval over which it increases

47. anonymous

right!

48. anonymous

and that's because of the domain that you were explaining before

49. anonymous

oops is said "increase" i meant "decrease"

50. anonymous

yea I figured!

51. anonymous

ok want to see a picture?

52. anonymous

sure

53. anonymous

http://www.wolframalpha.com/input/?i=x%2Bsqrt%281-x%29 click on "real valued plot" otherwise you will get complex plot which is not what you want

54. anonymous

cool, thanks

55. anonymous

yw $\frac{d}{dx}[\sqrt{x^2+2x-1}]=?$

56. anonymous

1/(2*sqrt(x^2+2x-1) * 2x+2)

57. anonymous

right, better knows as $\frac{x+1}{\sqrt{x^2+2x-1}}$ easy right?

58. anonymous