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Let kx6 be the space of polynomials in x of degree 6, and let U ‰ kx6 be the space of f(x) 2 kx6 such that the quintic equation f0(x) = 0 has no multiple roots It is obvious that U is a Zariski open dense subset of kx6 For f 2 U, we denote by Cf ‰ P2 the projective plane curve of degree 6 whose affine part is defined by y5 ¡f(x. C€xents Œˆ‚~ ' "ƒ7ƒ7 ƒ6a ‡¸lepos=0€ 4010 >DŽ ‰( w>‚üWhyÎotÂigÉdeasánd€sntervenAs?„‡„‡„‡„ height="2„ „ì0„ç„å. Solutions for MidTerm 1 ECE490 Introduction to Optimization Fall 18 1 Problem 1 (25 points) Show that the function f(x 1;x 2) = 8x 1 12x 2 x21 2x22 has only one stationary point, and that it is neither a local min or a local max, but a saddle.
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K o lp ml o o l k nk k m % p n s%m a o m p %o. May 22, 16 · Note that f(0)=(10)^k=1 Assuming k is a constant, we see that f'(x)=k(1x)^(k1) Thus f'(0)=k(10)^(k1)=k*1=k Using the point (0,1) and slope of k we can write the linearization function L(x)=k(x0)1=1kx. EQʏ cƳ g m1 ڶ d ?> 0Oښ ~ \V ٮ sZ' _G q O U { b 6 ky Z $ҳ ƻ ƅ!8Ρ ڮ aw 3 s nw !.
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