Is This the Correct Taylor Polynomial for sqrt(x) at x=100?

[Mathman23]

Hi

Given a function f(x) = sqrt(x) is the Taylor Polynomial of degree 2 for that function:

$$\frac{x^2}{2} - 99x + 4901$$ where x = 100 ?

Sincerely Fred

 

[Jameson]

Do you mean a=100, as in it's centered at a=100? If so, use Taylor's formula for approximating polynomials.

$$f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)*(x-a)^n}{n!}$$

I get the first two terms of the series are $$10+\frac{x-100}{20}$$

 

[HallsofIvy]

The Taylor's polynomial of degree 2 for a given function at a given point must match the function's value, first derivative and second derivative at that point.
Is $\sqrt{100}= 100^2/2- 99(100)+ 4901$?

Is the derivative of $\sqrt{x}$ at x= 100 equal to the derivative of that polynomial at x= 100?

Is the second derivative of $\sqrt{x}$ at x= 100 equal to the second derivative of that polynomial at x= 100?

If the answer to all three questions is correct, then that must be the
Taylor polynomial.