Find the coordinates of any stationary points of the curve y(x)=x^3-3x^2+3x+2

A stationary point is a point where the gradient of a curve is 0. The derivative of a curve gives us a function for the gradient at every point on the curve. So we have that dy(x)/dx=0 if and only if the curve is stationary at that point. The derivative of ax^n is nax^(n-1), so we find that dy/dx=3(x^2)-32x^1+3. There are only 3 terms as the derivative of 2 is 0, since 2 can be written as 2x^0 (any number to the power of 0 is 1), and 20*x^(-1)=0.We want to find the stationary points, so we are interested in where dy/dx=0. 0=3x^2-6x+3.This is just a quadratic equation, and can be solved as normal. To simplify, divide by 3,0=x^2-2x+1.Then we look for factorizations. We look for 2 numbers that add to make -2 and multiply to make 1. -1 and -1 satify this, so we can rewrite the quadratic as0=(x-1)(x-1)=(x-1)^2. this tells us when x=1, our equation is satisfied and the gradient is 0. Now we need to find out what y is when x =1. Putting x=1 into y(x) givesy(1)=(1)^3-3(1)^2+3(1)+2=1-3+3+2=3. So the coordinate of the only stationary point is (1,3)

RS
Answered by Robin S. Further Mathematics tutor

2815 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Given a^2 < 4 and a+2b = 8. Work out the range of possible values of b. Give your answer as an inequality.


Solve x^(-1/4) = 0.2


If z=4+i, what is 1/z? (in the form a+bi)


The equation of a curve is y = x^2 - 5x. Work out dy/dx


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences