When thinking about stationary points, it is important to remember that these points happen when the gradient of the curve is 0 (-> dy/dx = 0), or if you prefer, when the rate of change of a graph is 0.
So the first step is to differentiate the given equation:
If we take the first part of the equation, 6ln(x), we can split this into two seperate parts, ln(x) and 6, where these variables are multiplied together. Using the chain rule, where u = 6 and v = ln(x), we can calculate the differential by calculating v.du + u.dv, where dv is the differential of v, and du is the differential of u.
As the differential of ln(x) is 1/x, and the differential of 6 = 0, v.du is equal to ln(x).0, and u.dv is equal to 6.(1/x) which is equal to 6/x.
Next we move on to the differentiation of x2 , and as differentating involves multiplying by the current power, and then taking away 1 from the previous power, we get 2x.
Next, 8x differentiates to 8, as x1 multiplied by 1 and minus 1 is equal to x0 , and 8.1 = 8.
Finally, 3 differentiates to 0, as the derivative represents the rate of change, and a constant factor has a rate of change of 0 (as it is constant).
This leaves us with:
dy/dx = 6/x + 2x - 8
Getting rid of the fraction by multiplying by x, we get:
dy/dx = 6 + 2x2 - 8x (which rearranges to give)
-> dy/dx = 2x2 -8x + 6
By factorising the equation out, we get that x is either equal to 1 or 3:
= 2x2 - 8x + 6
= (2x - 6)(x-1)
-> 2x - 6 = 0 or x - 1 = 0
-> x = 3 or x = 1
Substitute in the values we have calculated into the original equation:
x = 1
y = 6ln(1) +12 - (8.1) + 3 (ln(1) = 0)
y = 1 - 8 + 3
y = -4
So when x = 1, y = -4
x = 3
y = 6ln(3) + 32 - (8.3) + 3
y = 6ln(3) + 9 - 24 + 3
y = 6ln(3) - 12
So when x = 3, y = 6ln(3) - 12
So the coordinates of the stationary points are:
(1, -4) and (3, 6ln(3) - 12)