# `f'''(x) = cos(x), f(0) = 1, f'(0) = 2, f''(0) = 3` Find `f`.

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### 1 Answer

`f'''(x)=cos(x)`

`f''(x)=intcos(x)dx`

`f''(x)=sin(x)+C_1`

Now let's find C_1 , given f''(0)=3

`f''(0)=3=sin(0)+C_1`

`3=0+C_1`

`C_1=3`

`:.f''(x)=sin(x)+3`

`f'(x)=int(sin(x)+3)dx`

`f'(x)=-cos(x)+3x+C_2`

Now let's find C_2 , given f'(0)=2

`f'(0)=2=-cos(0)+3(0)+C_2`

`2=-1+C_2`

`C_2=3`

`:.f'(x)=-cos(x)+3x+3`

`f(x)=int(-cos(x)+3x+3)dx`

`f(x)=-sin(x)+3(x^2/2)+3x+C_3`

Now let's find C_3 , given f(0)=1

`f(0)=1=-sin(0)+3(0^2/2)+3(0)+C_3`

`C_3=1`

`:.f(x)=-sin(x)+(3x^2)/2+3x+1`