Algebra, Limit and Derivative Problems

Date: 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
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