Question

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 :)

Answer 1

2nd derivative is?

Answer 2

i couldn't get that far :(

Answer 3

i hav the first deriv

Answer 4

we can go from there then, whats the 1st D ?

Answer 5

f'(x)=1(1/1+x^2)

Answer 6

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?

Answer 7

ohhhhhhhhhhhh quotient rule!!

Answer 8

\[y'=tan^{-1}(x)+(x+1)(1+x^2)^{-1}\]
\[y''=tan'^{-1}(x)+(x+1)'(1+x^2)^{-1}+(x+1)(1+x^2)'^{-1}\]
\[y''=(1+x^2)^{-1}+(1+x^2)^{-1}+(-2x)(x+1)(1+x^2)'^{-2}\]

Answer 9

quotient rule is fine; i like to use product on negative exponents to make life a bit easier to compute

Answer 10

i see

Answer 11

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"

Answer 12

0 : 0s and undefs
<--------------------->
- 1 + : curve directions

Answer 13

well, I got me + and - on the wrong sides, but still; its the point of inflection :)

Answer 14

so the answer is (-1, 0)?

Answer 15

nooo i meant (0,1)???

Answer 16

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

Answer 17

i have the option there, so it's either d or e?

Answer 18

correct, so when x=1

y=2 arctan(2)

Answer 19

so (1, pi/2)?

Answer 20

i got no idea what angle that is :) wolf to the rescue

Answer 21

arctan(1) right ....

Answer 22

lol

Answer 23

k

Answer 24

2*45^ = 2*pi/4 = pi/2 yep

Answer 25

thankss again :)