## Answers

f(x) = x^2 + 3x y = x^2 + 3x therefore dy = (2x + 3) dx Now x = 3 and dx = 0.1 therefore dy x = 3 dx = 0.1 = (2 times 3 + 3) times (0.1) = 0.9 f(x) = x^3 - 15 x^2 + 72 x therefore velocity d f(x)/dx = 3x^2 - 30x + 72 implies v = 3x^2 - 30x + 72 velocity at x = 5 seconds is 3 times 5^2 - 30 times 5 + 72 = -3 ft/sec Let at x seconds v = 0 implies 3x^2 - 15x + 72 = 0 implies x^2 - 5x + 24 = 0 implies x = 5 plusminus squareroot 25 - 4.1.24/2.1 = 5 plusminus squareroot -71/2, which is not real. Therefore The velocity will never be zero

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