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anonymous

  • 4 years ago

What are the coordinates of the inflection point on the graph of y=(x+1)arctanx a. (-1,0) b. (0,0) c. (0,1) d. (1, pi/4) e. (1, pi/2) work plzz :)

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  1. amistre64
    • 4 years ago
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    2nd derivative is?

  2. anonymous
    • 4 years ago
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    i couldn't get that far :(

  3. anonymous
    • 4 years ago
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    i hav the first deriv

  4. amistre64
    • 4 years ago
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    we can go from there then, whats the 1st D ?

  5. anonymous
    • 4 years ago
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    f'(x)=1(1/1+x^2)

  6. amistre64
    • 4 years ago
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    hmmm, lets see if I can verify that: \[y=(x+1)tan^{-1}(x)\] \[y'=(x+1)'tan^{-1}(x)+(x+1)tan'^{-1}(x)\] \[y'=tan^{-1}(x)+\frac{x+1}{1+x^2}\] \[y'=tan^{-1}(x)+(x+1)(1+x^2)^{-1}\] you agree?

  7. anonymous
    • 4 years ago
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    ohhhhhhhhhhhh quotient rule!!

  8. amistre64
    • 4 years ago
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    \[y'=tan^{-1}(x)+(x+1)(1+x^2)^{-1}\] \[y''=tan'^{-1}(x)+(x+1)'(1+x^2)^{-1}+(x+1)(1+x^2)'^{-1}\] \[y''=(1+x^2)^{-1}+(1+x^2)^{-1}+(-2x)(x+1)(1+x^2)'^{-2}\]

  9. amistre64
    • 4 years ago
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    quotient rule is fine; i like to use product on negative exponents to make life a bit easier to compute

  10. anonymous
    • 4 years ago
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    i see

  11. amistre64
    • 4 years ago
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    to clean it up we: \[y''=2(1+x^2)^{-1}+(-2x^2-2x)(1+x^2)^{-2}\] \[y''=(2(1+x^2)+(-2x^2-2x))(1+x^2)^{-2}\] \[y''=(2\ \cancel{+2x^2-2x^2}^{\ 0}-2x)(1+x^2)^{-2}\] \[y''=(2-2x)(1+x^2)^{-2}\] y'' = 0 when x=1; and is not undefined for any values so our only option is at x=1 to "test out"

  12. amistre64
    • 4 years ago
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    0 : 0s and undefs <---------------------> - 1 + : curve directions

  13. amistre64
    • 4 years ago
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    well, I got me + and - on the wrong sides, but still; its the point of inflection :)

  14. anonymous
    • 4 years ago
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    so the answer is (-1, 0)?

  15. anonymous
    • 4 years ago
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    nooo i meant (0,1)???

  16. amistre64
    • 4 years ago
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    http://www.wolframalpha.com/input/?i=2nd+derivative+%28x%2B1%29arctanx i dropped a negative someplace but still the same basic results so inflection is x=1 (x=1,y=) .... need to plug in x=1 into the y= f(x) part to determine its value unless we only have 1 option

  17. anonymous
    • 4 years ago
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    i have the option there, so it's either d or e?

  18. amistre64
    • 4 years ago
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    correct, so when x=1 y=2 arctan(2)

  19. anonymous
    • 4 years ago
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    so (1, pi/2)?

  20. amistre64
    • 4 years ago
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    |dw:1328031733509:dw|

  21. amistre64
    • 4 years ago
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    i got no idea what angle that is :) wolf to the rescue

  22. amistre64
    • 4 years ago
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    arctan(1) right ....

  23. anonymous
    • 4 years ago
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    lol

  24. anonymous
    • 4 years ago
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    k

  25. amistre64
    • 4 years ago
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    2*45^ = 2*pi/4 = pi/2 yep

  26. anonymous
    • 4 years ago
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    thankss again :)

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