Approximate  by linearizing f f(x)=3x^4-3x+2 a). Xo = 4 b). Xo=0

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We can approximate the value of a function near a point by using the tangent line at the point.

The tangent line for a differentiable function at the point c is given by:

`y=f(c)+f'(x)(x-c)`  and the limit as x approaches c is f(c).

Given `f(x)=3x^4-3x+2` we have `f'(x)=12x^3-3`

(1) For...

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We can approximate the value of a function near a point by using the tangent line at the point.

The tangent line for a differentiable function at the point c is given by:

`y=f(c)+f'(x)(x-c)`  and the limit as x approaches c is f(c).

Given `f(x)=3x^4-3x+2` we have `f'(x)=12x^3-3`

(1) For `x_0=4` we have the tangent line as:

`y=f(4)-f'(4)(x-4)`

`=758-765(x-4)`

`=-765x+3818`

So for values of x close to 4 the linear approximation is y=765x+3818

(2) For `x_0=0` we have the tangent line as:

`y=f(0)-f'(0)(x-0)`

`=2-(-3)(x)`

`=3x+2`

So for x values sufficiently close to 0 the approximation is y=3x+2

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