`f''(x) = 4 + 6x + 24x^2, f(0) = 3, f(1) = 10` Find `f`.

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Chapter 4, 4.9 - Problem 43 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f''(x)=4+6x+24x^2`

`f'(x)=int(4+6x+24x^2)dx`

`f'(x)=4x+(6x^2)/2+(24x^3)/3+C_1`

`f'(x)=4x+3x^2+8x^3+C_1`

`f(x)=int(4x+3x^2+8x^3+C_1)dx`

`f(x)=(4x^2)/2+(3x^3)/3+(8x^4)/4+C_1(x)+C_2`

`f(x)=2x^2+x^3+2x^4+C_1x+C_2`

Now let's find constants C_1 and C_2 , given f(0)=3 , f(1)=10

`f(0)=3=2(0)^2+0^3+2(0)^4+C_1(0)+C_2`

`C_2=3`

`f(1)=10=2(1)^2+1^3+2(1)^4+C_1+C_2`

`10=2+1+2+C_1+3`

`10=8+C_1`

`C_1=2`

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

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