At the point (2, 5) on the curve `y=x^3-2x+1` the gradient of the curve is increasing

At the point (2,3) on the curve y = x^(3) - 2x + 1 , the gradient of the curve increases

The points on the curve y=12x-x^3 at which the gradient is zero are

Find the point on the curve y=x^3+x^2+x at which the tangent to the curve is perpendicular to the x+y=1

If (-1,2) and and (2,4) are two points on the curve y=f(x) and if g(x) is the gradient of the curve at point (x,y) then the value of the integral int_(-1)^(2) g(x) dx is

If (-1,2) and and (2,4) are two points on the curve y=f(x) and if g(x) is the gradient of the curve at point (x,y) then the value of the integral int_(-1)^(2) g(x) dx is

Find the gradient of the curve y=sqrt(x^3) at x=4

find the co-ordinate of the point on the cute y = (x^2-1)/(x^2+1) (x>0) where the gradient of the tangent to the curve is maximum.

The normal at the point (1,1) on the curve 2y + x^2 = 3 is: