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Question

If f(x)=4x33x2+2x+k, where k is a constant to be determined and if f(0)=1,f(1)=4,findf(4)

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Solution

f(x)=4x33x2+2x+kf(x)dx=(4x33x2+2x+k)dxf(x)=x4x3+x2+kx+cf(0)=04+03+02+k(0)+c=1c=1f(1)=14+13+12+k+1=4k=0f(x)=x4+x3+x2+1f(4)=44+43+42+1256+64+16+1=337

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