The calculator will find the quadratic approximation to the given function at the given point, with steps shown.

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

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