y=x^5-3 find dy/dx

Question

y=x^5-3

1- find dy/dx

2- find the co-ordinates of any stationary points and determine their nature

3- sketch the curve

Answer 1

1. dy/dx=5x^4
2. Stationary point is where dy/dx=0, so x=0 and y=-3, the point (0,-3).
3. d2y/dx2=20x^3 which is 0 when x=0 so it's a saddle point where the curve is flat at the stationary point and, if we put x slightly less than zero (say x=-0.1) then y=-10^-5-3 which is slightly smaller (more negative) than -3, and dy/dx is small and positive, so the gradient or slope is positive; and if we put x=0.1, y is slightly bigger than -3 but the slope is still small and positive, we can see that at the stationary point the curve is flat, and near this point it is almost flat with a positive gradient (/). The graph of the curve starts at a large negative value when x is large and negative and as x approaches zero and y approaches -3 from the left it dramatically bends till it's almost flat. To the right of -3 it straightens out and continues upward, cutting the x axis when x^5=3 at about x=1.25 (5th root of 3). The curve is clearly very narrow with a largely positive slope at all times.