# Given that f(x) is differentiable over the real set, what are a and b? f(x)=2x^3+ax^2+b, if x=<1 f(x)=x^2-2bx+3a, if x>1

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Since the function is differentiable over the real set of numbers, then the function is continuous in x=1.

Therefore, left limit of f(x), x->1 with x<1=right limit of f(x), x->1, x>1=f(1).

lim (2x^3+ax^2+b) = lim (x^2-2bx+3a)

We'll substitute x by 1 both sides:

2 + a + b = 1 - 2b + 3a

We'll move all terms in a and b to the left side and the numbers alone to the right.

a - 3a + b + 2b = 1 - 2

-2a + 3b = -1

2a - 3b = 1 (1)

Since the function is continuous in x = 1, then it is differentiable in x = 1.

f'(x) = 6x^2 + 2ax, x<1

f'(x) = 2x - 2b, x>1

lim f'(x) (left) = lim f'(x) (right)

lim (6x^2 + 2ax) = lim (2x - 2b)

We'll substitute x by 1 both sides:

6+2a = 2-2b

2a+2b=-4

a+b = -2 (2)

We'll solve the system of equations (1) and (2):

2a - 3b = 1

a + b = -2

2a - 3b + 3a + 3b = 1 - 6

5a = -5 => a = -1

-1 + b = -2

b = -1

**The given function is differentiable over R set, if and only if a = -1 and b = -1.**