since the curve y=ax3+bx+4 passes through (2,14)therefore, 14=a(2)3+b(2)+4 5=4a+b.....(1) The slope of tangent to the curve y=ax3+bx+4 at point(2,14) is,
dy
dx
=3ax2+b (
dy
dx
)(2,14)=3a(2)2+b 21=12a+b.....(2) On solving equation (1) and (2), we get a=2 b=−3