# find a linearization function and approximate a) sqrt(25.07) and b) sqrt(24.96) the linearization function is= a) the approximation of sqrt(25.07) is= b) the approximation sqrt(24.96) is=

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Let we have a function `f(x)=x^(1/2)` ,which is to be linearized.

We use Taylor's Theorm for linearization.

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

`f(x)=x^(1/2)`

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

`a.` x=25.07

`x_0=25`

`f(25.07)=f(25)+(25.07-25)(1/2)f'(25)`

`=(25)^(1/2)+.07xx(1/2)(25)^(-1/2)`

`=5+.07xx(1/2)(1/5)=5+.007=5.007`

b. x=24.96

`x_0=25`

`f(24.96)=f(25)+(24.96-25)(1/2)f'(25)`

`=5-.04(1/2)(1/5)=5-.004=4.996`