Find the turning points of the curve y = 3x^4 - 8x^3 -3
Differentiate to get:dy/dx = 12x^3 -24x^2Factorise and set equal to 0:12x^2(x-2) = 0Solve to get x = 2 and x =0.
DC
Related Maths A Level answers
All answers ▸differentiate y=(4x^3)-5/x^2
Let y = 4t/(t^2 + 5). Find dy/dt, writing your answer in it's simplest form, and find all values of t for which dy/dt = 0
What's the gradient of the curve y=x^3+2x^2 at the point where x=2?
How would you work out the equation of the normal at a point (2,5) given the equation of a line?