Please help me out with this one

Find A, B and C so that the function that is defined by

f(x) =

2 + (2/pi) * arctan 1/(x+1) when x < −1
3x + C|x| when −1 ≤ x ≤ 1
1 / (Ax+B) when x > 1 ,

will be continuous at x = −1 and derivable at x = 1.

Please help me solve the value of A B and C, since im stuck and really need help. Step by step if possible please!

f(x) is continuous at x=-1, so its left-hand limit must be equal to its right hand limit at x=-1

left-hand limit = lim [x--->-1] 2 + (2/pi) * arctan 1/(x+1)
= 2 + (2/pi)*(pi/2) = 2+1 = 3
right-hand limit = lim [x--->-1] 3x + C|x| = -3 + C

==> -3 +C = 3 ==> C= 6

f(x) is derivable at x=1, so f(x) is continuous (i.e. left-hand limit = right-hand limit at x=1)
and its rigth derivative must equal its left derivative at x=1

left-hand limit = lim [x-->1] 3x + 6|x| = 3+6 = 9
right-hand limit = lim [x-->1] 1/(Ax+B) = 1/(A+B)

==> 1/(A+B) = 9 or A+B = 1/9.....(1)

at x=1
left derivative = 3 + 6 =9
(because for x>0, |x| = x)
right derivative = -A/(A+B)^2

==> -A/(A+B)^2 = 9

-A[1/(A+B)]^2 = 9

from (1), we know that 1/(A+B) = 9
==> -81A =9
==> A= -1/9
B= 1/9 - A = 2/9