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find a linearization function and approximate a.(100.06)^1/3 and b.(999.93)^1/3 the...

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rmunoz90 | Student, Undergraduate | Salutatorian

Posted May 15, 2013 at 7:15 PM via web

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find a linearization function and approximate a.(100.06)^1/3 and b.(999.93)^1/3

the linearization function is T(x)=

a. the approximation of (100.06)^1/3 is=

b. the approximation (999.93)^1/3 is=

Tagged with calculus, math

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 15, 2013 at 10:58 PM (Answer #1)

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The function that we wish to linearize is:

`f(x) =x^(1/3)`

And its derivative is:

`f'(x)=1/3x^(-2/3)`

The linearization of a function at x=a is defined as:

`T(x) = f(a) + f'(a)(x-a)` 

`T(x)=a^(1/3)+1/3a^(-2/3)(x-a)`

For 100.06, x=100.06 and a=100:

`T(100.06)=100^(1/3)+1/3(100^(-2/3))(100.06-100)`

Solving this equation we find that the linear approximation for the cubed root of 100.06 is approximately 4.6425.

For 999.93, x=999.93 and a=999

`T(999.93)=999^(1/3)+1/3(999^(-2/3))(999.93-999)`

Solving this equation we find that the linear approximation for the cubed root of 999.93 is approximately 9.9997

Sources:

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oldnick | Valedictorian

Posted May 16, 2013 at 1:40 AM (Answer #2)

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`f(x)=x^(1/3)`      `f'(x)=1/3 x^(-2/3)=1/3xx 1/(root(3)(x^2))`

Set  `x_0= 100`   so :  `f(100)=root(3)(100)`  and `f'(100)=1/30 1/root(3)(10)` Then:

 

`f(100.06)-f(100)=1/30 xx 1/root(3)(10) xx 6/100` `=1/500 xx1/root(3)(10)`

`f(100.06)= 1/500 xx1/root(3)(10) + root(3)(100)` `=1/500 xx1/root(3)(10) xx5001 `

`f(100.06)=4.6425171513795014481885583661932`

 

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

`f(1000)=10`   `f('1000)=1/3 xx 1/100`

`f(999.93)=10-1/3 xx 1/100 xx 7/100=` `10-7/30.000`

`f(999.993)=9.9997stackrel(-)6`

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