A curve has equation: y = x^3 - 3x^2 + 5. Show that the curve has a minimum point when x = 2.

A minimum point will have a gradient of 0 (although so will a maximum point or a point of inflection). dy/dx = 3x2-6x. We can substitute x = 2 into this equation to give 0 (alternatively solve 3x2-6x = 0 to give x = 2 as a root). Thus the gradient at x = 2 is 0 so it is either a minimum point, maximum point or point of inflection. To prove it is in fact a minimum point, we can differentiate the original equation of the curve once more to give d2y/dx2= 6x - 6 (the second derivative). Substituting x = 2 into this equation gives a positive value. Positive value = minimum point; negative value = maximum point; zero value requires further investigation by finding gradients of x values slightly smaller and larger than 2 (this is also a possible alternative method to determine which type of stationary point it is to begin with).

Related Further Mathematics GCSE answers

All answers ▸

The equation 3x^2 – 5x + 4 = 0 has roots P and Q, find a quadratic equation with the roots (P + 1/2Q) and (Q + 1/2P)


What is the distance between two points with x-coordinates 4 and 8 on the straight line with the equation y=(3/4)x-2


Factorise the following quadratic x^2 - 8 + 16


Find the definite integral of f(x) = 12/(x^2+10x+21) with limits [-1,1]. Give your answer to 2 decimal places.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences