TEST2PracSoln - SampleT2Solns.nb 1 Solutions to Practice...

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Solutions to Practice Problems for Test 2, Math 2015 1) Ÿ - 1 5 H 3 f H x L - g H x LL x = 3 Ÿ - 1 5 f H x L x - Ÿ - 1 5 g H x L x = 3 I Ÿ - 1 2 f H x L x + Ÿ 2 5 f H x L x M - I Ÿ - 1 2 g H x L x + Ÿ 2 5 g H x L x M = 3 H 3 - 1 L - H 5 + 2 L =- 1 There is no rule that would allow us to calculate Ÿ - 1 2 f H x L g H x L x . 2) M £ = 28 - 7 M can be rewritten as M £ =- 7 H M - 4 L , so the equilibrium solution is M = 4 and the general solution is M = 4 + Ce - 7 t . Solving M (0)=5 gives 5 = 4+ C , or C = 1. So the solution is M = 4 + e - 7 t . 3) F is increasing where f is positive, so from about x = 0 to x ª 1.6. F is then decreasing for x greater than about 1.6. The largest value for F occurs at the local maximum at x ª 1.6, and the smallest value is at the end of the interval (after F has been decreasing) at x = 4. A graph of the antiderivative F is shown below: 1.5 2 2.5 3 3.5 4 -6 -4 -2 4) a) Avg Value = 1 ÅÅÅÅÅÅÅÅ 6 - 0 Ÿ 0 6 1 ÅÅÅÅ 3 H 30
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TEST2PracSoln - SampleT2Solns.nb 1 Solutions to Practice...

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