courtrigrad
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Hello all
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
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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