`f''(x) = -2 + 12x - 12x^2, f(0) = 4, f'(0) = 12` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 39 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f''(x)=-2+12x-12x^2`

`f'(x)=int(-2+12x-12x^2)dx`

`f'(x)=-2x+12(x^2/2)-12(x^3/3)+c_1`

`f'(x)=-2x+6x^2-4x^3+c_1`

Now let's find constant c_1 , given f'(0)=12

`f'(0)=12=-2(0)+6(0^2)-4(0^3)+c_1`

`c_1=12`

`f'(x)=-2x+6x^2-4x^3+12`

`f(x)=int(-2x+6x^2-4x^3+12)dx`

`f(x)=-2(x^2/2)+6(x^3/3)-4(x^4/4)+12x+c_2`

`f(x)=-x^2+2x^3-x^4+12x+c_2`

Now let's find constant c_2 , given f(0)=4

`f(0)=4=-0^2+2(0)^3-0^4+12(0)+c_2`

`c_2=4`

`:.f(x)=-x^2+2x^3-x^4+12x+4`

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