How do I find the turning point/stationary point of this equation y = 9(x − 1)^2 − (x − 1)^4?

Differentiate and equate to zero 

y = 9(x − 1)^2 − (x − 1)^4

dy/dx = 2(9)(x-1)^1(1) - 4(x - 1)^3(1) = 0 

Tidying up 

18x - 18 -4(x^3 - 3x^2 + 3x - 1) = 0 

18x - 18 - 4x^3 + 12x^2 - 12x + 4 = 0 

-4x^3 + 12x^2 + 6x -14  = 0 

By 'trial and error

x = 1 

Substitute 

-4 + 12 + 6 - 14 = 0 

So when x = 1 As the stationary/turning point , then y = 0 

So the co-ords are (1,0) 

Another way is to expand the brackets, collect 'like terms' and then differentiate and equate to zero. 

y = 9(x^2 - 2x + 1) - ( x^4- 4x^3 + 6x^2 - 4x + 1)

y = 9x^2 - 18x + 9 - x^4 + 4x^3 - 6x^2 + 4x - 1 

y = -x^4 +  4x^3 + 3x^2 - 14x + 8 

dy/dx = -4x^3 + 12x^2 + 6x - 14 = 0 

Then as above. 

NB By 'Trial and error' you have to try and find a number for 'x' that makes the differentiated eq'n equal to zero. To do this try 0, 1, -1, 2, -2, et seq.  

The differentiated equation does not readily factorise. 

However, you can divide the differentiated equation by. 'x - 1', the factor from above. The result will be a quadratic eq'n which may factorise to find the other x-values of the turning points.  The basic equation is raised to the power of '4' , so there will be three(3) turning points. 

y = 9(x - 1)² - (x - 1)⁴

so, dy/dx = 18(x - 1) - 4(x - 1)³

Now, stationary/turning points are when dy/dx = 0

i.e. when 18(x - 1) - 4(x - 1)³ = 0

or, 2(x - 1)[9 - 2(x - 1)²] = 0

Then, 2(x - 1) = 0 or 9 - 2(x - 1)² = 0

i.e. x - 1 = 0 or (x - 1)² = 9/2

so, x = 1, x = 1 ± (3√2/2)

or, x = 1, x = (2 ± 3√2)/2

i.e. x = 1, x = (2 + 3√2)/2 and x = (2 - 3√2)/2

A sketch of the function is below.

:)>  

Here's a trick.

Consider the function 

w = 9x^2 - x^4.

dw/dx = 18x - 4x^3.

The denominator is 0 when

x = 0, and also when x = +/- (3/2)*sqrt(2)..  

So the three stationary points of w are at x = 0, +(3/2)sqrt(2) and -(3/2)sqrt(2).

Are they turning points?  For that we need

d2w/dx2 = 18 - 12x^2.  

The stationary points must also be turning points, as d2w/dx2 is not 0 there.

Then it should be plan that the three turning points of y are at

x = 1, x = 1 + (3/2)sqrt(2), and x = 1 - (3/2)sqrt(2).