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This is my question,

A function of 2 variable is given by,

$f(x,y) = e^{2x-3y}$

How to find tangent approximation to $f(0.244, 1.273)$ near $(0,0)?$

I need some guidance for this question. Am i suppose to do the linear approximation or quadratic approximation?

Need some explanation for the formula. Thanks

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    I've never heard of the word tangent used outside of lines, planes, or any "flat" surfaces2012-09-25

2 Answers 2

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Linear(Tangent) approximation of $f(x,y)$ about $(a,b)$ is given by,

$f(x,y)\approx f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$

In your problem $f(x,y)=e^{2x-3y}$

$f_x(x,y)=2e^{2x-3y}$ and $f_y(x,y)=-3e^{2x-3y}$

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    @David: exactly :)2012-09-25
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More precise approximation we obtain if represent $f(x,\,y)$ as $f(x,\,y)=e^{2x-3y}=e^3e^{2x-3y-3}=e^3e^{2x-3(y-1)}.$ Then apply formula for tangent approximation to function $g(x,\,y)=e^{2x-3(y-1)}$ with $a=0; \,b=1.$