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

Related Maths GCSE answers

All answers ▸

if x^2 + 9x + 20 = 0, what are the possible values of x?


Expand and simplify (x+6)(x-6)


Fred is reshaping his garden. He decides to make into a right-angled triangle ABC. The length AB= 5m and the length BC= 12 What is the length of AC?


Solve the linear equation 12x - 4 = 3x + 2