Stationary points help C1
The question told me to "Find the coordinates for the stationary points on the curve y=x^3-3x^2+4" But I wasn't too confident on if I was correct.
Firstly I diffrentiated it so that I get 3x^2-6x.
I then tried to find a way to factorise it and got 2x(x-2)
I then concluded that x=2 but I felt like I should of got two x values. I think I'm going completely off.
Can someone tell me if I made a mistake when doing this?
Firstly I diffrentiated it so that I get 3x^2-6x.
I then tried to find a way to factorise it and got 2x(x-2)
I then concluded that x=2 but I felt like I should of got two x values. I think I'm going completely off.
Can someone tell me if I made a mistake when doing this?
Original post by Fruitbasket786
The question told me to "Find the coordinates for the stationary points on the curve y=x^3-3x^2+4" But I wasn't too confident on if I was correct.
Firstly I diffrentiated it so that I get 3x^2-6x.
I then tried to find a way to factorise it and got 2x(x-2)
I then concluded that x=2 but I felt like I should of got two x values. I think I'm going completely off.
Can someone tell me if I made a mistake when doing this?
Firstly I diffrentiated it so that I get 3x^2-6x.
I then tried to find a way to factorise it and got 2x(x-2)
I then concluded that x=2 but I felt like I should of got two x values. I think I'm going completely off.
Can someone tell me if I made a mistake when doing this?
Stationary points mean the gradient = 0
3x^2-6x=0
Factorise it (check your factorisation, it's wrong)
Then you do have 2 solutions for x
3x=0 or x-2=0
Posted from TSR Mobile
Original post by gdunne42
Stationary points mean the gradient = 0
3x^2-6x=0
Factorise it (check your factorisation, it's wrong)
Then you do have 2 solutions for x
3x=0 or x-2=0
Posted from TSR Mobile
3x^2-6x=0
Factorise it (check your factorisation, it's wrong)
Then you do have 2 solutions for x
3x=0 or x-2=0
Posted from TSR Mobile
I see, I accidentally added 2x instead of 3x. Thanks!
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