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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    |dw:1439260678254:dw|

  2. anonymous
    • one year ago
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    step one find the derivative

  3. anonymous
    • one year ago
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    |dw:1439260757824:dw|

  4. anonymous
    • one year ago
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    let me know when you get \[f'(x)=1-\frac{1}{2\sqrt{1-x}}\]

  5. anonymous
    • one year ago
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    I'm just having trouble trying to solve for x

  6. anonymous
    • one year ago
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    ok lets back up a second

  7. anonymous
    • one year ago
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    exponential notation is great for computing certain derivatives, but it totally sucks for doing actual numeric computation

  8. anonymous
    • one year ago
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    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
    • one year ago
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    and via the chain rule \[\frac{d}{dx}[\sqrt{f(x)}]=\frac{f'(x)}{2\sqrt{f(x)}}\]

  10. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yeah the power rule works for sure, but the derivative never changes

  13. anonymous
    • one year ago
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    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
    • one year ago
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    |dw:1439261266064:dw|So would the function look like this

  15. anonymous
    • one year ago
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    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
    • one year ago
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    yeah what you wrote now we can do it

  17. anonymous
    • one year ago
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    Do we multiple the 1 by the denominator of the other fraction?

  18. anonymous
    • one year ago
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    you want to set it equal to zero and solve probably a good first step

  19. anonymous
    • one year ago
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    oh kk

  20. anonymous
    • one year ago
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    notice that the deriviative has domain \(x<1\) which is not surprising because the original function has domain \(x\leq 1\)

  21. anonymous
    • one year ago
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    also notice that the derivative is strictly decreasing

  22. anonymous
    • one year ago
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    So does x equal 3/4 ?

  23. anonymous
    • one year ago
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    which means all that is left to do is solve \[1-\frac{1}{2\sqrt{x-1}}=0\]

  24. anonymous
    • one year ago
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    yes it does

  25. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yeah you can do that if it is not obvious

  28. anonymous
    • one year ago
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    then translate to increasing decreasing for \(f\) and you are done

  29. anonymous
    • one year ago
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    quick quiz \[\frac{d}{dx}[\sqrt{1-x^2}]=?\]

  30. anonymous
    • one year ago
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    so (-infinity, 3/4) is incr and decreasing from (3/4, infinity)

  31. anonymous
    • one year ago
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    oh no hold the phone

  32. anonymous
    • one year ago
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    increasing on \((-\infty, \frac{3}{4})\) is right

  33. anonymous
    • one year ago
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    1/2*sqrt(1-x^2) *-2

  34. anonymous
    • one year ago
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    1/(2*sqrt(1-x^2))*-2

  35. anonymous
    • one year ago
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    close the numerator should be \(-2x\)

  36. anonymous
    • one year ago
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    and of course the two's cancel

  37. anonymous
    • one year ago
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    Oh yea, was in a rush

  38. anonymous
    • one year ago
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    ok lets get back to your question

  39. anonymous
    • one year ago
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    \[f(x)=x+\sqrt{1-x}\]right?

  40. anonymous
    • one year ago
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    is it undefined from 3/4 to infinity

  41. anonymous
    • one year ago
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    don't write that is increasing on \((\frac{3}{4},\infty)\) or your teacher will think you are daft

  42. anonymous
    • one year ago
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    the domain is \[(-\infty, 1]\] right?

  43. anonymous
    • one year ago
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    right so we just dont include the 3/4 to infinity part, right?!

  44. anonymous
    • one year ago
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    so it is only increasing on \[(\frac{3}{4},1)\]

  45. anonymous
    • one year ago
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    yeah you include it , but dont to to infinity

  46. anonymous
    • one year ago
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    three fourth is less than one, so there is an interval over which it increases

  47. anonymous
    • one year ago
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    right!

  48. anonymous
    • one year ago
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    and that's because of the domain that you were explaining before

  49. anonymous
    • one year ago
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    oops is said "increase" i meant "decrease"

  50. anonymous
    • one year ago
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    yea I figured!

  51. anonymous
    • one year ago
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    ok want to see a picture?

  52. anonymous
    • one year ago
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    sure

  53. anonymous
    • one year ago
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    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
    • one year ago
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    cool, thanks

  55. anonymous
    • one year ago
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    yw \[\frac{d}{dx}[\sqrt{x^2+2x-1}]=?\]

  56. anonymous
    • one year ago
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    1/(2*sqrt(x^2+2x-1) * 2x+2)

  57. anonymous
    • one year ago
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    right, better knows as \[\frac{x+1}{\sqrt{x^2+2x-1}}\] easy right?

  58. anonymous
    • one year ago
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    yea, not bad

  59. anonymous
    • one year ago
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    so no more screwing around with rational exponents when dealing with square roots!!

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