What are a,b,c if f(x)=x^3+ax^2+bx+c, if x=1?f is differentiated two times on real set number

Since f is differentiable, we'll differentiate with
respect to x, to determine the 1st derivative.

f'(x) = 3x^2
+ 2ax + b, if x<1

f'(x) = 1/[1+(x-1)^2], if
x>1

We'll differentiate again, to determine the 2nd
derivative:

f"(x) = 6x+  2a,
x<1

f"(x) = -[1 +
(x-1)^2]'/[1+(x-1)^2]^2

f"(x) = 2(x-1)/[1+(x-1)^2]^2,
x>1

Since only a continuous function could be
differentiated, we'll impose the continuity
constraints:

lateral limits for x->1 = the value of
the function for x = 1

lim f(x) = f(1) <=> 1
+ a + b + c = arctan(1-1) = arctan 0 = 0 (1)

If f could be
differentiated 2 times, then lim f'(x)(x<1) =
limf'(x)(x>1)

3 + 2a + b =
1/[1+(1-1)^2]

3 + 2a + b = 1
(2)

Also lim f''(x)(x<1) =
limf''(x)(x>1):

6 + 2a =
2(1-1)/[1+(1-1)^2]^2

6 + 2a =
0

2a = -6

a =
-3

We'll substitute a in
(2):

3 + 2a + b = 1 <=> 3 - 6 + b = 1
=> b = 4

1 + a + b + c =
0

c = -1 - a - b

c = -1 + 3 -
4

c = -2

The
requested values of a,b,c are: a = -3 , b = 4 and c =
-2.