The curve C has equation y = 3x^4 – 8x^3 – 3 Find (i) dy/dx (ii) the co-ordinates of the stationary point(s)

i) dy/dx=12x^3-24x^2ii) the stationary points occur when dy/dx = 0 so we must find the solutions to 12x^3-24x^2=0.12x^3-24x^2= 12x^2(x-2)=0Therefore our stationary points are when 12x^2=0 ie x=0 and x-2=0 ie x=2.Substituting our x co-ordinates into the original equation, we get our co-ordinates out as (0,-3) and (2,-19)

Related Maths A Level answers

All answers ▸

3. The point P lies on the curve with equation y=ln(x/3) The x-coordinate of P is 3. Find an equation of the normal to the curve at the point P in the form y = ax + b, where a and b are constants.


If y = 5x^3 - 2x^2 + 2, what is dy/dx?


Find dy/dx for y=5x^3-2x^2+7x-15


What is the y-coordinate minimum point of y = 3x^2 + x - 4