Algebra, Limit and Derivative ProblemsDate: 7/28/96 at 1:52:32 From: Suryadie Gemilang Subject: Algebra, Limit and Derivative Problems 1. Let x+y=1 and x^3+y^3=19. What is the value of x^2 + y^2? 2. f(x)=sqrt(2x+4). lim f(h)-f(0)/h = h->0 3. d(8x^3+8x^2+6x+1)/d(2x) = Date: 7/28/96 at 12:53:20 From: Doctor Anthony Subject: Re: Algebra, Limit and Derivative Problems (1) You can factorize x^3 + y^3 to (x+y)(x^2 - xy + y^2) = 19 and since x+y = 1 we have x^2 - xy + y^2 = 19 2x^2 + 2y^2 - 2xy = 38 ....(1) Then also (x+y)^2 = x^2 + y^2 + 2xy = 1 ....(2) Adding equations (1) and (2) 3x^2 + 3y^2 = 39 so x^2 + y^2 = 13. (2) This is simply the value of dy/dx at x=0. y = sqrt(2x+4) = (2x+4)^(1/2) dy/dx = (1/2)(2x+4)^(-1/2)*2 = 1/{(2x+4)^(1/2)} and when x=0 this gives dy/dx = 1/2 (3) This can be written (1/2)d{8x^3 + 8x^2 + 6x + 1}/dx = (1/2){24x^2 + 16x + 6} = 12x^2 + 8x + 3 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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