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

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.