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If we have [itex] f'(x) = x^{-\frac{1}{3}} - 1 [/itex] and [itex] f(8) = 4 [/itex] find [itex] f(x) [/itex]. Ok so [itex] f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + C [/itex]. Since [itex] f(8) = 4 [/itex] then [itex] f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + 6 [/itex]

If we have [itex] f''(x) = 2x^{2}, f'(3) = 10, f(3) = 6 [/itex] find [itex] f(x) [/itex]. So we have [itex] f'(x) = \frac{2}{3}x^{3} - 8 [/itex]. Now [itex] f(x) = \frac{1}{6}x^{4} - 8x + C [/itex] So would [itex] f(x) = \frac{1}{6}x^{4} - 8x + 16.5 [/itex]?

Thanks

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# Homework Help: Verification of Problems

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