y=20x-x^2-2x^3. Curve has a stationary point at the point M where x=-2. Find the x coordinate of the other stationary point of the curve and the value of the second derivative of both of these point, hence determining their nature.

Differentiate to get dy/dx=20-2x-6x^2Then stationary points occur when dy/dx = 0 so 0 = 20-2x-6x^2 Factorise to get x= -2, x=5/3Differentiate dy/dx to get second derivative = -2-12x at x=5/3 is -22 so max pointat x=-2 second derivative is 24>0 so min point.

EJ

Related Maths A Level answers

All answers ▸

A curve has the equation y=sin(x)cos(x), find the gradient of this curve when x = pi. (4 marks)


A curve has parametric equations x=t(t-1), y=4t/(1-t). The point S on the curve has parameter t=-1. Show that the tangent to the curve at S has equation x+3y+4=0.


What is the indefinite integral of (x^4)*(-sin(x)) dx


Solve the following equations. Leave answers in simplest terms a)e^(3x-9)=8. b) ln(2y+5)=2+ln(4-y)