If a curve has equation y=(4/3)x^3-2x^2-24x+16, find dy/dx and find the coordinates of the turning points.

y=(4/3)x3-2x2-24x+16Step 1: Understand the questiondy/dx means differentiate the function of y with respect to xturning points are where the gradient of the function changes and will be found by setting dy/dx = 0 [note dy/dx = 0 is not always a turning point]Step 2: Solve the problemdy/dx = 4x2-4x-24simplifies to: dy/dx = x2-x-6now to find turning points: set dy/dx=0 such that x2-x-6=0 which factorises out as (x+2)(x-3)=0Thus, the roots to the equation are x=-2 and x=3Then to find coordinates, sub the x values back into equation to find their corresponding y valuesThus, final solution is (-2, 136/3) and (3,-38)Step 3: Reflect and consolidate learningUnderstand what you have solved - you have found local maxima and minima pointsPotential further qs: is a stationary point necessarily a turning point?, how can you show that these points are turning points?

JP
Answered by Johann P. Maths tutor

3004 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

what is differentiation for?


Solve the simultaneous equations: 5x + y = 21 and x - 3y = 9


A particle is moving along a straight line. The fixed point O lies on this line. The displacement of the particle from O at time t seconds is s metres where s = 2t3 – 12t2 + 7t(a) Find an expression for the velocity, v m/s, of the particle at time t.


Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y - 3x = 13


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences