Answer to Question #94765 in Calculus for Unknown287159

Question

Find the co-ordinates of the points on the curve Y=X^3-2x^2+3x-5 where its gradient is 2

Answer 1

$\frac{dy}{dx}=3x^2-4x+3.$

$\frac{dy}{dx}=2, so\; 3x^2-4x+3=2 \; or\; 3x^2-4x+1=0.$

Thus, $x=\frac{1}{3}\;or\;x=1.$

$y(\frac{1}{3})=-\frac{113}{27}, y(1)=-3.$

Points on the curve where its gradient is 2: $(\frac{1}{3},-\frac{113}{27}) \; and\; (1,-3).$