Suppose f(x)=18x+4e2x\displaystyle {f}{''}{\left({x}\right)}=-{18}{x}+{4}{e}^{{-{2}{x}}}, f(0)=7\displaystyle {f}'{\left({0}\right)}={7}, and f(0)=5\displaystyle {f{{\left({0}\right)}}}={5}.

Then f(x)=\displaystyle {f{{\left({x}\right)}}}=  

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