Consider the (real-valued) function f: R²R defined by 0 for (x, y) = (0,0), for (x, y) = (0,0). f(x,y)= 2-3 x² + y² ar a

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answerhappygod
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Consider the (real-valued) function f: R²R defined by 0 for (x, y) = (0,0), for (x, y) = (0,0). f(x,y)= 2-3 x² + y² ar a

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Consider The Real Valued Function F R R Defined By 0 For X Y 0 0 For X Y 0 0 F X Y 2 3 X Y Ar A 1
Consider The Real Valued Function F R R Defined By 0 For X Y 0 0 For X Y 0 0 F X Y 2 3 X Y Ar A 1 (12.34 KiB) Viewed 33 times
Consider The Real Valued Function F R R Defined By 0 For X Y 0 0 For X Y 0 0 F X Y 2 3 X Y Ar A 2
Consider The Real Valued Function F R R Defined By 0 For X Y 0 0 For X Y 0 0 F X Y 2 3 X Y Ar A 2 (32.98 KiB) Viewed 33 times
Consider the (real-valued) function f: R²R defined by 0 for (x, y) = (0,0), for (x, y) = (0,0). f(x,y)= 2-3 x² + y² ar ac

(c) Let y: R→ R2 be a differentiable function [that is, y is a differentiable curve in the plane R2] which is such that 7(0) = (0,0), and y(t) (0,0) whenever y(t) = (0,0) for some t E R. Now, set g(t):= f(y(t)) (the composition of f and y), and prove that (this real- valued function of one real variable) g is differentiable at every t E R. Also prove that if y E C¹ (R, R²), then g = C¹(R. R). [Note that this shows that f has "some sort of derivative" (i.e., some rate of change) at the origin whenever it is restricted to a smooth curve that goes through the origin (0,0).] (d) In spite of all this, prove that f is not (Fréchet) differentiable at the origin (0,0). (Hint: Show that the formula (D.f) (0,0) = ((V)(0.0), v) fails for some direc- tion(s) v. Here (.) denotes the standard dot product in the plane R².)
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