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).

TC
Answered by Tom C. Further Mathematics tutor

8480 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Let y = (4x^2 + 3)^4. Find dy/dx.


How can you divide an algebraic expression by another algebraic expression?


How would you differentiate x^x?


Find the tangent to the equation y=x^2 -2x +4 when x=2


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning