What are a,b,c if f(x)=x^3+ax^2+bx+c, if x<1 or f(x)=tan^-1(x-1) if x>=1?

f is differentiated two times on real set number

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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.**

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