You are Here: Home >< Maths

# Integration watch

1. f)(i) Show that the curve y = x3 - 6x2 + 11x - 6 cuts the x-axis at x = 1, 2 and 3.

(ii) Find the x-coordinates of the two stationary points on the curve.

(iii) Find d2y/dx2 and establish which is a maximum and which is a minimum.

i) (x3 - 6x2 + 11x - 6) / (x-1) = (x^2 - 5x + 6)

x^2 - 5x + 6 = (x-2)(x-3)

I got this bit correct... but(ii)

dy/dx = 3x^2 - 12x + 11

x 33 || +(-12) -> none go in so use other method

x = -b +- √b^2 - 4ac
------------------
2a

a = 3
b = -12
c = 11

x = 12 +- √144 - 132
---------------------
6
x = 2+- √12
x = 2+- 3√2
It says its wrong
A: 2+-√3/3 .... what did I do wrong?

I would of wrote
2+ 3√2(minimum)
2- 3√2(maximum)
A:
2-√3/3(maximum)
2+√3/3(minimum)
2. (Original post by ckfeister)

x = 12 +- √144 - 132
---------------------
6
x = 2+- √12
x = 2+- 3√2
It says its wrong
A: 2+-√3/3 .... what did I do wrong?

That's fine.

Then, here's the simplification in detail.

3. (Original post by ckfeister)
f)(i) Show that the curve y = x3 - 6x2 + 11x - 6 cuts the x-axis at x = 1, 2 and 3.

(ii) Find the x-coordinates of the two stationary points on the curve.

(iii) Find d2y/dx2 and establish which is a maximum and which is a minimum.

i) (x3 - 6x2 + 11x - 6) / (x-1) = (x^2 - 5x + 6)

x^2 - 5x + 6 = (x-2)(x-3)

I got this bit correct... but(ii)

dy/dx = 3x^2 - 12x + 11

x 33 || +(-12) -> none go in so use other method

x = -b +- √b^2 - 4ac
------------------
2a

a = 3
b = -12
c = 11

x = 12 +- √144 - 132
---------------------
6
x = 2+- √12
x = 2+- 3√2
It says its wrong
A: 2+-√3/3 .... what did I do wrong?

I would of wrote
2+ 3√2(minimum)
2- 3√2(maximum)
A:
2-√3/3(maximum)
2+√3/3(minimum)
For the first part, the first thing that went wrong is that as to have suggested. Also in your answer you should have but you divided the 12 by 6 but forgot to divide the by 6.
For the second part remember that If at a stationary point then it is a maximum point and the opposite is true for if it is a minimum point.
4. (Original post by ghostwalker)

That's fine.

Then, here's the simplification in detail.

How do you get the signs up you got?
5. (Original post by ckfeister)
How do you get the signs up you got?
LaTex - there's a link in the Useful Resources widget.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 26, 2016
The home of Results and Clearing

### 2,791

people online now

### 1,567,000

students helped last year
Today on TSR

### Took GCSEs this summer?

Fill in our short survey for Amazon vouchers!

### University open days

1. Bournemouth University
Wed, 22 Aug '18
2. University of Buckingham
Thu, 23 Aug '18
3. University of Glasgow
Tue, 28 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams