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+ For the function F(x) = 24 – 14.2 + 60.rº – 70x find minimum value using a. Newtons method starting with initial point of

tolerance of 0.001

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# Given that FIX) = 24-147 +6012-704 So Let Now flon = 8 428-14x*+1202 -70 ne know at extrenum Fle) = 0 P(x) Flen) = 473-142?F0)>0 at 2 = 0.620 so FEM) has a minimum at 0.620 ५ i Menütmiem vahe F(0.620) 0.62099-1460.620) +6010-62032 -70(0.620) = 0.1

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