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I am confused as to how to proceed with the following linear approximation:

$$ \frac{(2.01)^2}{\sqrt{.95}} $$

I know that we need to define a function such that $f(x) = \frac{(2.01)^2}{\sqrt{.95}}$ and then Taylor expand around some value that we can compute exactly, but I do not know how to define an appropriate function to do this. Any hints?

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set $f(x)=\frac{(2+x)^2}{\sqrt{1-5x}}$ expand it then set $x=0.01$.
using linear approximation $ $ $f(x)=4+14x$ $ $ $f(0.01)=4.14$.

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You may also do it using the linear approximation of $f(x,y) = \frac{x^2}{\sqrt{y}}$ at $(x_0,y_0)= (2,1)$ with $\Delta x = 0.01$ and $\Delta y = 0.05$:

$$f(x,y) \stackrel{linear}{\approx} f(x_0,y_0) + \frac{\partial f}{\partial x}(x_0,y_0)\Delta x + \frac{\partial f}{\partial y}(x_0,y_0)\Delta y$$ With

  • $\frac{\partial f}{\partial x} = \frac{2x}{\sqrt{y}}$
  • $\frac{\partial f}{\partial y} = -\frac{x^2}{2\sqrt{y^3}}$

you get $$\frac{(2.01)^2}{\sqrt{.95}} \approx 4 + 4\cdot 0.01 - 2\cdot 0.05 =3.94$$

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