Question

The curve y = aX^3– 2X^2– X + 7 has a gradient of 3 at the point where X =3.

Determine the value of a.

Answer 1

Gradient: $y'(x)=3ax^2-4x-1.$

$y'(3)=3\to3a*3^2-4*3-1=3\to27a=16\to a=\frac{16}{27}.$