Sujit Kumar
Last Activity: 6 Years ago
f(x)=a0+a1x+a2x2+a3x3+.....anxn
f(0)=a0 also given.... f(0)=2
Therefore a0=2
f’(x)=a1+2a2x+3a3x2+......nanxn-1
Therefore f’(0)=a1
also given.... f’(0)=3
Therefore a1=3
Now you can double differentiate f(x)=f’’(x) and get the values of a2=1 and a3=0.5 and input them in the original equation f(x) and now a2 and a3 have to be equal to a4 and a5
You can go on until you get a pattern.