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設f(x)=x^m 帶入點(2,√2)得:√2=2^m解得:m=1/2即:f(x)=x^(1/2)__(也就是根號X) 設g(x)=x^n 帶入點(-2,4)得:4=(-2)^n解得:n=2即:g(x)=x^2 (2)由√x0 得x1
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