Question

What are the local extrema, if any, of `f (x) =2ln(x^2+3)-x`?

Answer 1

`f(x) = 2ln(x^2+3) -x` has a local minimum for `x=1` and a local maximum for `x=3`

Explanation:

We have:

`f(x) = 2ln(x^2+3) -x`

the function is defined in all of `RR` as `x^2+3 > 0 AA x`

We can identify the critical points by finding where the first derivative equals zero:

`f'(x) = (4x)/(x^2+3)-1= - (x^2-4x+3)/(x^2+3)`

`- (x^2-4x+3)/(x^2+3) = 0`

`x^2-4x+3 = 0`

`x= 2+-sqrt(4-3)=2+-1`

so the critical points are:

`x_1 = 1` and `x_2 = 3`

Since the denominator is always positive, the sign of `f'(x)` is the opposite of the sign of the numerator `(x^2-4x+3)`

Now we know that a second order polynomial with positive leading coefficient is positive outside the interval comprised between the roots and negative in the interval between the roots, so that:

`f'(x) < 0` for `x in (-oo, 1)` and `x in (3,+oo)`
`f'(x) > 0` for `x in (1,3)`

We have then that `f(x)` is decreasing in `(-oo, 1)`, increasing in `(1,3)`, and again decreasing in `(3,+oo)`, so that `x_1 = 1` must be a local minimum and `x_2=3` must be a local maximum.

graph{2ln(x^2+3) -x [-1.42, 8.58, -0.08, 4.92]}