Question

What are the critical numbers of `f(x) = x (5 + 6 ln x)`?

Answer 1

There is one critical point at `(e^(-11/6), - e^(-11/6))`

Explanation:

You need to use the product rule; `d/dx(uv)=u(dv)/dx+v(du)/dx`

So with `f(x)=x(5+6lnx)` we have;

`f'(x) = { (x)(d/dx(5+6lnx)) + (5+6lnx)(d/dxx) }`
`:. f'(x) = { (x)(6/x) + (5+6lnx)(1) }`
`:. f'(x) = 6 + 5+6lnx`
`:. f'(x) = 11+6lnx`

At critical points, we have `f'(x)=0`

`f'(x)=0 => 11 + 6lnx = 0`
`:. lnx = -11/6`
`:. x = e^(-11/6) (~~ 0.16)`

When `x = e^(-11/6) => f(x) = e^(-11/6)(5+6(-11/6))`
:. `f(x) = e^(-11/6)(5-6) = - e^(-11/6)`

So there is one critical point at `(e^(-11/6), - e^(-11/6))`

graph{x(5+6lnx) [-1, 2, -2, 5]}