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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 :)
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i couldn't get that far :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i hav the first deriv

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we can go from there then, whats the 1st D ?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0hmmm, 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
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhhhhhhhhhh quotient rule!!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[y'=tan^{1}(x)+(x+1)(1+x^2)^{1}\] \[y''=tan'^{1}(x)+(x+1)'(1+x^2)^{1}+(x+1)(1+x^2)'^{1}\] \[y''=(1+x^2)^{1}+(1+x^2)^{1}+(2x)(x+1)(1+x^2)'^{2}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0quotient rule is fine; i like to use product on negative exponents to make life a bit easier to compute

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0to clean it up we: \[y''=2(1+x^2)^{1}+(2x^22x)(1+x^2)^{2}\] \[y''=(2(1+x^2)+(2x^22x))(1+x^2)^{2}\] \[y''=(2\ \cancel{+2x^22x^2}^{\ 0}2x)(1+x^2)^{2}\] \[y''=(22x)(1+x^2)^{2}\] y'' = 0 when x=1; and is not undefined for any values so our only option is at x=1 to "test out"

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.00 : 0s and undefs <>  1 + : curve directions

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0well, I got me + and  on the wrong sides, but still; its the point of inflection :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so the answer is (1, 0)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nooo i meant (0,1)???

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0http://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
 4 years ago
Best ResponseYou've already chosen the best response.0i have the option there, so it's either d or e?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0correct, so when x=1 y=2 arctan(2)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328031733509:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i got no idea what angle that is :) wolf to the rescue

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.02*45^ = 2*pi/4 = pi/2 yep
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