# How do you find f given #f''(x)=-2+36x-12x^2# and f(0)=2, f'(0)=1?

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To find ( f(x) ) given ( f''(x) = -2 + 36x - 12x^2 ) and ( f(0) = 2, f'(0) = 1 ), follow these steps:

- Integrate ( f''(x) ) to find ( f'(x) ).
- Integrate ( f'(x) ) to find ( f(x) ).
- Apply initial conditions ( f(0) = 2 ) and ( f'(0) = 1 ) to solve for the constants of integration.

Step 1: [ \int f''(x) ,dx = \int (-2 + 36x - 12x^2) ,dx ] [ f'(x) = -2x + 18x^2 - 4x^3 + C_1 ]

Step 2: [ \int f'(x) ,dx = \int (-2x + 18x^2 - 4x^3 + C_1) ,dx ] [ f(x) = -x^2 + 6x^3 - x^4 + C_1x + C_2 ]

Step 3: Apply initial conditions: [ f(0) = 2 \Rightarrow 2 = 0 + 0 + 0 + C_2 ] [ f'(0) = 1 \Rightarrow 1 = 0 + 0 + 0 + C_1 ]

Therefore, ( C_1 = 1 ) and ( C_2 = 2 ).

So, the solution is: [ f(x) = -x^2 + 6x^3 - x^4 + x + 2 ]

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