courtrigrad
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Hello all
If we have f'(x) = x^{-\frac{1}{3}} - 1 and f(8) = 4 find f(x). Ok so f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + C. Since f(8) = 4 then f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + 6
If we have f''(x) = 2x^{2}, f'(3) = 10, f(3) = 6 find f(x). So we have f'(x) = \frac{2}{3}x^{3} - 8. Now f(x) = \frac{1}{6}x^{4} - 8x + C So would f(x) = \frac{1}{6}x^{4} - 8x + 16.5?
Thanks
If we have f'(x) = x^{-\frac{1}{3}} - 1 and f(8) = 4 find f(x). Ok so f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + C. Since f(8) = 4 then f(x) = \frac{3}{2}x^{\frac{2}{3}} -x + 6
If we have f''(x) = 2x^{2}, f'(3) = 10, f(3) = 6 find f(x). So we have f'(x) = \frac{2}{3}x^{3} - 8. Now f(x) = \frac{1}{6}x^{4} - 8x + C So would f(x) = \frac{1}{6}x^{4} - 8x + 16.5?
Thanks

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