If a curve has equation y = (-8/3)x^3 - 2x^2 + 4x + 18, find the two x coordinates of the stationary points of this curve.

The first step in solving this question is to differentiate equation y with respect to x. To differentiate a given 'x' term in an equation use the following method --> xn becomes nx(n-1).y = (-8/3)x3 - 2x2 + 4x + 18dy/dx = -8x2 - 4x + 4A stationary point is found when dy/dx is equal to zero. This means dy/dx can now be solved for x using the quadratic formula. x = (- b +/- √(b2 - 4ac))/2aUsing this formula and identifying a = -8, b = -4 and c = 4, x = (4 +/- √(16 + 128))/ - 16 = (4 +/- 12)/ - 16the two x coordinates of the stationary points of the curve y can be found to be:x = (4 + 12)/ - 16 = -1and x = (4 - 12)/ - 16 = 1/2

RT
Answered by Robyn T. Maths tutor

3093 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

f(x)=(2x+1)/(x-1) with domain x>3. (a)Find the inverse of f(x). (b)Find the range of f(x). (c) g(x)=x+5 for all x. Find the value of x such that fg(x)=3.


Solve the simultaneous equations: y = x - 2 and y^2 + x^2 = 10


Solve the equation: log5 (4x+3)−log5 (x−1)=2.


The height (h) of water flowing out of a tank decreases at a rate proportional to the square root of the height of water still in the tank. If h=9 at t=0 and h=4 at t=5, what is the water’s height at t=15? What is the physical interpretation of this?


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