anonymous
  • anonymous
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 :)
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
2nd derivative is?
anonymous
  • anonymous
i couldn't get that far :(
anonymous
  • anonymous
i hav the first deriv

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amistre64
  • amistre64
we can go from there then, whats the 1st D ?
anonymous
  • anonymous
f'(x)=1(1/1+x^2)
amistre64
  • amistre64
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?
anonymous
  • anonymous
ohhhhhhhhhhhh quotient rule!!
amistre64
  • amistre64
\[y'=tan^{-1}(x)+(x+1)(1+x^2)^{-1}\] \[y''=tan'^{-1}(x)+(x+1)'(1+x^2)^{-1}+(x+1)(1+x^2)'^{-1}\] \[y''=(1+x^2)^{-1}+(1+x^2)^{-1}+(-2x)(x+1)(1+x^2)'^{-2}\]
amistre64
  • amistre64
quotient rule is fine; i like to use product on negative exponents to make life a bit easier to compute
anonymous
  • anonymous
i see
amistre64
  • amistre64
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"
amistre64
  • amistre64
0 : 0s and undefs <---------------------> - 1 + : curve directions
amistre64
  • amistre64
well, I got me + and - on the wrong sides, but still; its the point of inflection :)
anonymous
  • anonymous
so the answer is (-1, 0)?
anonymous
  • anonymous
nooo i meant (0,1)???
amistre64
  • amistre64
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
anonymous
  • anonymous
i have the option there, so it's either d or e?
amistre64
  • amistre64
correct, so when x=1 y=2 arctan(2)
anonymous
  • anonymous
so (1, pi/2)?
amistre64
  • amistre64
|dw:1328031733509:dw|
amistre64
  • amistre64
i got no idea what angle that is :) wolf to the rescue
amistre64
  • amistre64
arctan(1) right ....
anonymous
  • anonymous
lol
anonymous
  • anonymous
k
amistre64
  • amistre64
2*45^ = 2*pi/4 = pi/2 yep
anonymous
  • anonymous
thankss again :)

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