# Stationary pointsWatch

#1
I need help on this question, don't know where they get (0,2) from.

I have worked out that dy/dx = 96x + 128x^3=0

Here is the question and mark scheme:

It's part c

Posted from TSR Mobile
0
4 years ago
#2
(Original post by Jimmy20002012)
I need help on this question, don't know where they get (0,2) from.

I have worked out that dy/dx = 96x + 128x^3=0

Here is the question and mark scheme:

It's part c

Posted from TSR Mobile
Put the value x=0 you found back into the equation to obtain the y value.
0
4 years ago
#3
The stationary point has x-coordinate 0; 0 is the only value of x that solves the equation 96x + 128x^3 = 0. Get the y-coordinate by using x = 0: y = (1 + 2(0))^4 + (1 - 2(0))^4
0
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