f.4differentiation

-3-

2010-03-31 2:57 am

1) Let y=x^3+(k-1)x^2+3x+2,where k is a constant. If the equation dy/dx =0 has
Only one real solution,find the possible values of k.

The answer is 4 or -2. Please show clear steps.

2) Let y =(2x^2+7x)^1/2 and z^2lnz=4x-2
a) find dy/dx
bi) find the value of x when z =1
bii)hence,or otherwise find dy/dx when z=1.

The answers: a)4x+7/2(2x^2+7x)^1/2
bi1/2 bii) 9/16
I want to ask part bii only

pls show steps.thx

更新1:

where does this 4(k-1)^2-36=0 come from?

更新2:

sorry !bii)should be hence,or otherwise find dy/dz when z=1.

回答 (1)

myisland8132

2010-03-31 3:12 am

✔ 最佳答案

1 dy/dx=3x^2+2(k-1)x+3
Since determinant=0=>4(k-1)^2-36=0
That is k-1=3 or -3
k=4 or -2
2(a) dy/dx=(4x+7)/[2√ (2x^2+7x)]
(b)(i) when z=1=>z^2lnz=0

4x-2=0=>x=1/2
(ii) Sub.x=1/2=>dy/dx=9/4

Your anser of b(ii) is wrong

2010-03-30 19:53:53 補充：
z^2lnz=4x-2
[z+(lnz)(2z)]dz=4dx
when z=1,dz=4dx=>dx/dz=1/4

dy/dz=(dy/dx)(dx/dz)=9/16

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