`f''(x) = 2x^3 + 3x^2 - 4x + 5, f(0) = 2, f(1) = 0` Find `f`.

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Chapter 4, Review - Problem 72 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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`f''(x) = 2x^3 + 3x^2 - 4x + 5`

`or, f'(x) = {(1/2)x^4} + x^3 - 2x^2 + 5x + a`

`or, f(x) = {(1/10)*x^5} + (x^4/4) - {(2/3)x^3} + {(5/2)x^2} + ax + b`

`Now, f(0) = 2`

`i.e. 2 = b`

`Also, f(1) = 0`

`i.e. 0 = (1/10) + (1/4) - (2/3) + (5/2) + a + 2`

`or, a = -131/60`

`Thus, f(x) = {(1/10)*x^5} + (x^4/4) - {(2/3)x^3} + {(5/2)x^2} - {(131/60)x} + 2`

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