A curve has equation y = (x-1)e^(-3x). The curve has a stationary point M. Show that the x-coordinate of M is 4/3.

(In the following answer, y' denotes dy/dx, i.e. the derivative of y with respect to x, where y is a function of x)

y = (x-1)e^-3x = uv, where u = x-1 and v = e^-3x.
We have that u' = 1 and using the Chain Rule, v' = e^-3x * (-3) = -3e^-3x.
(For further clarification on acquiring v' if needed, we let z = -3x, then v = e^z and z' = -3, since the Chain Rule states that d/dx( g(f(x)) ) = g'(f(x)) * f'(x), we get that v' = (e^z)' = e^z * z' = e^z * (-3) = -3e^-3x ).


Using the Product Rule ( y' = (uv)' = u'v + uv') we get,
y' = 1 * e^-3x + (x-1)*(-3e^-3x), rearranging we then get,
= e^-3x - 3e^-3x(x-1), factorising e^-3x gives,
= e^-3x(1 - 3(x - 1)), expanding brackets leads to,
= e^-3x(1 - 3x + 3), collecting terms results in,
= e^-3x(4 - 3x).


We get a stationary point when y' = 0.
Thus at the x-coordinate of M, m, we have,
y'(m) = e^-3m(4 - 3m) = 0, since e^-3m > 0 for all real numbers m, we have,
4 - 3m = 0, and so
m = 4/3.