In this question we will calculate the Taylor Polynomial for $f(x) = \sqrt{x} + 2$ about $x = 7$.
The formula for the Taylor Polynomial of degree 3 for the function $f(x)$ about $x = a$ is:
$T_3(x) = f(a) + f'(a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3$
In this case, that means we need to find $f(7)$, $f'(7)$, $f''(7)$ and $f'''(7)$.
$f(x) = \sqrt{x} + 2$, so $f(7) =
$
f'(x) =
, so f'(7) =
f''(x) =
, so f''(7) =
f'''(x) =
, so f'''(7) =
Therefore the Taylor Polynomial for $f(x) = \sqrt{x} + 2$ about $x = 7$ is:
+
$(x - 7) +
(x - 7)^2 +
(x - 7)^3$
