How do you find the stationary points of the curve with equation y=4x^3-12x+1
The stationary points are located where the gradient of the function is equal to 0. The first derivative of the function describes the gradient so we must differentiate the function to find dy/dx.
Using differentiation rules, we find that dy/dx = 12x^2 - 12.
Now we must set the derivative to zero and solve for x:
dy/dx = 12x^2-12 = 0 12(x^2-1) = 012(x-1)(x+1) = 0
This has solutions x=1 and x=-1. Therefore the curve has 2 stationary points; one with x-coordinate 1 and one with x coordinate -1.
To find the corresponding y-coordinates we substitute these x values into the equation of the curve y=4x^3-12x+1.
When x=1, y = 4*(1)^3 - 121+1 = 4 - 12 +1 = -7.
When x=-1, y = 4(-1)^3 - 12*(-1) + 1 = -4 + 12 + 1 = 9.
Hence the stationary points of the curve with the equation y=4x^3-12x+1 are (1, -7) and (-1, 9).
Related Maths A Level answers
All answers ▸How do you find the point or points of intersection of a straight line and a circle?
Consider a cone of vertical height H (in metres) and base radius R (in metres) which is full with water. The cone, at time t=0, starts to leak such that it loses water at a rate of k m^3 per second. Give an expression for the rate of change of H.
How do you differentiate 5x
How do I deal with parametric equations? x = 4 cos ( t + pi/6), y = 2 sin t, Show that x + y = 2sqrt(3) cos t.