Let y = f ( x ) = ( x 1 2 + 2 x 2 , x 1 x 2 − 3 x 1 )Is the linear approximation just f ( y ) = f ( x ) + A ( y − x ) whenever y is approximately near x?

Name: Let y = f ( x ) = ( x 1 2 + 2 x 2 , x 1 x 2 − 3 x 1 )Is the linear approximation just f ( y ) = f ( x ) + A ( y − x ) whenever y is approximately near x?
Uploaded: 2022-05-06T15:56:33+00:00
Description: Let y = f ( x ) = ( x 1 2 + 2 x 2 , x 1 x 2 − 3 x 1 )Is the linear approximation just f ( y ) = f ( x ) + A ( y − x ) whenever y is approximately near x?

Question

Let   y  =  f  (  x  )  =  (      x    1    2    +  2      x    2    ,      x    1        x    2    −  3      x    1    )Is the linear approximation just   f  (  y  )  =  f  (  x  )  +  A  (  y  −  x  ) whenever y is approximately near   x?

Wade Atkinson · Accepted Answer

A differentiable function   f  :            R        2    →            R        2   can be approximated by it&#039;s derivative in the sense that given a point, say,   (  1  ,  1  ), you can write:  f  (  1  +  δ  x  ,  1  +  δ  y  )  ≈  f  (  1  ,  1  )  +  D  f  (  1  ,  1  )  (  δ  x  ,  δ  y  )where   D  f  (  1  ,  1  ) is a matrix applied to the vector   (  δ  x  ,  δ  y  ). So, yes, if   y is near   x (as (  (  1  +  δ  x  ,  1  +  δ  y  )) is near (  (  1  ,  1  ))), then you can approximate the difference   f  (  y  )  −  f  (  x  ) by the derivative evaluated at   x, applied to the difference   y  −  x. That is:  f  (  y  )  ≈  f  (  x  )  +  D  f  (  x  )  (  y  −  x  )Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:  f  (  y  )  ≈  f  (  x  )  +      f    ′    (  x  )  (  y  −  x  )

Let y = f ( x ) = ( <msubsup> x 1 2 </msubsup> +

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