**Taylor series** is an example of infinite series derived from the expansion of `f(x)` about a single point. It is represented by infinite sum of `f^n(x)` centered at `x=c.` The **general formula for Taylor series** is:

`f(x) = sum_(n=0)^oo (f^n(c))/(n!) (x-c)^n`

or

`f(x) = f(c) + f'(c) (x-c)+ (f'(c))/(2!) (x-c)^2+ (f'(c))/(3!) (x-c)^3+ (f'(c))/(4!) (x-c)^4+...`

To evaluate the given function `f(x) =sqrt(x)` , we may express it in terms of fractional exponent. The function becomes:

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

Apply the definition of the Taylor series by listing the `f^n(x) ` up to `n=3.`

We determine each derivative using Power Rule for differentiation: `d/(dx) x^n = n*x^(n-1)` .

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

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

`= 1/2x^(-1/2) or1/(2x^(1/2) )`

`f^2(x) = d/(dx) (1/2x^(-1/2))`

`= 1/2 * d/(dx) (x^(-1/2))`

`= 1/2*(-1/2x^(-1/2-1))`

`= -1/4 x^(-3/2) or -1/(4x^(3/2))`

`f^3(x) = d/(dx) (-1/4x^(-3/2))`

`= -1/4 *d/(dx) (x^(-3/2))`

`= -1/4*(-3/2x^(-3/2-1))`

`= 3/8 x^(-5/2) or 3/(8x^(5/2))`

Plug-in `x=4` , we get:

`f(x) = (4)^(1/2)`

`= 2`

`f'(4)=1/(2*4^(1/2))`

`=1/(2*2)`

`=1/4`

`f^2(4)=-1/(4*2^(3/2))`

`= -1/(4*8)`

` = -1/32`

`f^3(4)=3/(8*4^(5/2))`

`= 3/(8*32)`

`= 3/256`

Applying the formula for Taylor series centered at `c=4` , we get:

`sum_(n=0)^3 (f^n(4))/(n!)(x-4)^n`

` =f(4) + f'(4) (x-4)+ (f'(4))/(2!) (x-4)^2+ (f'(4))/(3!) (x-4)^3`

` =2+ (1/4) (x-4)+ (-1/32)/(2!) (x-4)^2+ (3/256)/(3!) (x-4)^3 `

` =2+ (1/4) (x-4)+ (-1/32)/(2!) (x-4)^2+ (3/256)/(3!) (x-4)^3 `

` =2+ 1/4 (x-4)-1/(32*2) (x-4)^2+ 3/(256*6) (x-4)^3 `

`=2+ 1/4 (x-4)-1/64 (x-4)^2+ 3/1536 (x-4)^3`

`=2+ 1/4 (x-4)-1/64 (x-4)^2+ 1/512 (x-4)^3 `

The **Taylor polynomial **of degree `n=3` for the given function `f(x)=sqrt(x)` centered at ` c=4` will be:

`P(x) =2+ 1/4 (x-4)-1/64 (x-4)^2+ 1/512 (x-4)^3 `

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