# 二次近似计算器

Enter a function:
Enter a point:
x_0=

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

## Solution

Your input: find the quadratic approximation to $f(x)=\sqrt{x} + \frac{5}{\sqrt{x}}$ at $x_0=9$.

A quadratic approximation is given by $Q(x)\approx f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{1}{2}f^{\prime \prime}(x_0)(x-x_0)^2$.

We are given that $x_0=9$.

Firstly, find the value of the function at the given point: $y_0=f(x_0)=\frac{14}{3}$.

Secondly, find the derivative of the function, evaluated at the point: $f^{\prime}\left(9\right)$.

Find the derivative: $f^{\prime}\left(x\right)=\frac{x - 5}{2 x^{\frac{3}{2}}}$ (steps can be seen here).

Next, evaluate the derivative at the given point.

$f^{\prime}\left(9\right)=\frac{2}{27}$.

Now, find the second derivative of the function evaluated at the point: $f^{\prime \prime}\left(9\right)$.

Find the second derivative: $f^{\prime \prime}\left(x\right)=\frac{15 - x}{4 x^{\frac{5}{2}}}$ (steps can be seen here).

Next, evaluate the second derivative at the given point.

$f^{\prime \prime}\left(9\right)=\frac{1}{162}$.

Plugging the found values, we get that $Q(x)\approx \frac{14}{3}+\frac{2}{27}\left(x-\left(9\right)\right)+\frac{1}{2}\left(\frac{1}{162}\right)\left(x-\left(9\right)\right)^2$.

Simplify: $Q(x)\approx \frac{x^{2}}{324} + \frac{x}{54} + \frac{17}{4}$.

Answer: $Q(x)\approx \frac{x^{2}}{324} + \frac{x}{54} + \frac{17}{4}$.