If the function f(x) = ax^3 + bx^2 + 1 has a point of inflection at (-1,2).

At the point of inflection the second derivative is equal to 0.

f(x) = ax^3 + bx^2 + 1

=> f'(x) = 3ax^2 + 2bx

=> f''(x) = 6ax + 2b

6ax +...

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If the function f(x) = ax^3 + bx^2 + 1 has a point of inflection at (-1,2).

At the point of inflection the second derivative is equal to 0.

f(x) = ax^3 + bx^2 + 1

=> f'(x) = 3ax^2 + 2bx

=> f''(x) = 6ax + 2b

6ax + 2b = 0

As the point of inflection is at (-1 , 2)

6a*(-1) + 2b = 0

=> -6a + 2b = 0

=> 3a = b

Also, f(-1) = 2

=> a(-1) + 3a(1) + 1 = 2

=> -a + 3a + 1 = 2

=> 2a = 1

=> a = 1/2

b = 3a = 3/2

**The value of a = 1/2 and b = 3/2**