If g(t)=1t4+3t2+9\displaystyle {g{{\left({t}\right)}}}={1}{t}^{{4}}+{3}{t}^{{2}}+{9} find
g(0)=\displaystyle {g{{\left({0}\right)}}}=  
g′(0)=\displaystyle {g}'{\left({0}\right)}=  
g′′(0)=\displaystyle {g}{''}{\left({0}\right)}=  
g′′′(0)=\displaystyle {g}{'''}{\left({0}\right)}=  
g(4)(0)=\displaystyle {{g}^{{{\left({4}\right)}}}{\left({0}\right)}}=  
g(5)(0)=\displaystyle {{g}^{{{\left({5}\right)}}}{\left({0}\right)}}=  

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