`f''(x) = 1 - 6x + 48x^2, f(0) = 1, f'(0) = 2` Find `f`.

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Chapter 4, Review - Problem 71 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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`f''(x) = 1 - 6x + 48x^2`

`f'(x) = x - 3x^2 + 16x^3 + a`

`Now, f'(0) = 2`

`i.e. 2 = a`

`Thus, f'(x) = x - 3x^2 + 16x^3 + 2`

`f(x) = (x^2/2) - x^3 + 4x^4 + 2x + b`

`Now, f(0) = 1`

`Thus, b = 1`

`Hence, f(x) = (x^2/2) - x^3 + 4x^4 + 2x + 1`

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