Question

If $f^{{}^{\prime }}\left(x\right)=4x^3-3x^2+2x+k$, where k is a constant to be determined and if $f\left(0\right)=1,f\left(1\right)=4,findf\left(4\right)$

Answer 1

$f^{\prime }\left(x\right)=4x^3-3x^2+2x+k\int f^{\prime }\left(x\right)dx=\int (4x^3-3x^2+2x+k)dxf\left(x\right)=x^4-x^3+x^2+kx+cf\left(0\right)=0^4+0^3+0^2+k\left(0\right)+c=1⟹c=1f\left(1\right)=1^4+1^3+1^2+k+1=4⟹k=0f\left(x\right)=x^4+x^3+x^2+1f\left(4\right)=4^4+4^3+4^2+1256+64+16+1=337$