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find the linearization function and approximate a) cube(1000.06) and b) cube(999.93)...

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elpony | Student, Undergraduate | eNoter

Posted May 16, 2013 at 10:50 PM via web

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find the linearization function and approximate a) cube(1000.06) and b) cube(999.93)

The linearization function is T(x)=

a) the approximation of cube(1000.06) is=

b) the approximation of cube(999.93) is=

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pramodpandey | College Teacher | Valedictorian

Posted May 17, 2013 at 4:24 AM (Answer #1)

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Let we have a function `f(x)=x^3` ,which is to be linearized.

We use Taylors, theorem for this,

`f(x)=f(x_0+x-x_0)=f(x_0)+(x-x_0)f'(x_0)`

This linearlized function of f(x) .

`f(x)=x^3`

`f'(x)=3x^2`

`a.`  x=1000.06

`x_0=1000`

`f(1000.06)=f(1000)+(1000.06-1000)(3f'(1000))`

`=(10^3)^3+.06(3)(10^3)^2`

`=10^9+.18xx10^6`

`=10^6(1000+.18)`

`=1000.18xx10^6`

`b.` x=999.93

`x_0=1000`

`f(999.93)=f(1000)+(999.93-1000)(3f'(1000))`

`=(10^3)^3-.07(3xx(10^3)^2)`

`=10^9-.07xx3xx10^6`

`=10^6xx(1000-.21)`

`=999.79xx10^6`

``

 

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