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  1. NicolasPan

    The Function h(x)=e^(2x)-2

    1.a) Dh=ℝ b)The solution can go through this path: (e^x)^2-e^x-2=0 If e^x=y i) then the equation is transformed like that: y^2-y-2=0 then the solutions are y=(1±3)/2 and from i) we get e^x=(1±3)/2 BUT since the function e^x can only give positive results for x∈ℝ the only acceptable solution...
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