# linearize f(x)=6x^2-5x+6 near a. x0=8 b.x0=3 a. y= b. y=

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(b) Now linearization of `f(x)=6x^2-5x+6` near `x_0=3` .

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

`f(3)=45` .

`f'(3)=31.`

So, `L(3)=45+(x-3)31`

or, `L(3)=31x-48.`

This is the required linearization of f(x) near x_0=3.

Given function is f(x)=`6x^2-5x+6` .

To linearize a function f(x) near a point `x=x_0` , we symply give the Taylor's expansion of the first order about `x=x_0.`

i.e. `L(x_0)=f(x_0)+(x-x_0)f'(x_0)` where `f'=d/dx f(x)`

Now (a) we want to linearize the given function f(x) near `x_0=8` .

`f(8)=6.8^2-5.8+6=350.`

`f'(x)=d/dx(6x^2-5x+6)=12x-5`

so, `f'(8)=12.8-5=91` .

Now Linearization of f(x) near `x_0=8` is given by

`L(8)=350+(x-8)91`

`L(8)=91x-728`

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