How do you find the turning point for y=(2x-1)^3+1

How did the tp become (1/2, 1) Please explain
Also how do you find the tp for this as well y= -3x^3 +1
Please explain properly thanks

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Thanks!

I think you mean the point of inflection. If so, then what you do is take the second derivative of the equation, set it to equal zero, and solve for X.

For example: derivative of (2X - 1)^3 + 1 is 24X^2 - 24X + 6
The derivative of that is 48X - 24 so when you set that to zero you get 24 = 48X then X = .5 or 1/2

As for the second one, the first derivative is -9X^2, then the second derivative is -18X, so when set to zero, X = 0!

Then for both of those, just substitute the X value into the original equation to get the Y value! Hope this helps!

y' = 3(2x-1)^2 (2)
y'' = 12(2x-1)(2) = 0
x = 1/2
y = f(1/2) = 1

In the same way, you can get the tp for y= -3x^3 +1 is (0, 1)