Consider this multivariable function. f(x,y)=xy+2x+y−36 a) What is the value of f(2,−3)? b) Find all x-values such that f(x,x)=0
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a) We find f(2,−3) be replacing x by 2 and by -3 in f(x,y) f(2,−3)=2(−3)+2(2)+(−3)−36 f(2,−3)=−6+4−3−36=−41 b) Now, we solve f(x,x)=0 as follow. f(x,x)=0 x2+2x+x−36=0 x2+3x−36=0 x=−b±b2−4ac2a=(−3±32−4(1)(−36)))2(1)=−3±1532
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