I have the following Open Loop Transfer Function:
$$H_{ol}s = \frac{k(s+4)}{(s+1)(s+2)(s+3)}$$
To find the breakaway point of the unmatched poles we find the characteristic equation of the closed loop transfer function, rearrange to make $k$ the subject of the expression, set the derivative $\frac{dk}{ds} = 0$, find the roots, and one of these roots should lie within the root locus.
the characteristic equation is as follows:
$$CE: (s+1)(s+2)(s+3) +k(s+4) = 0$$
Making $k$ the subject shows,
$$k = -\frac{(s+1)(s+2)(s+3)}{(s+4)}$$
$\frac{dk}{ds} = 0$ should then be,
$$2s^3 +18s^2+48s+38 = 0$$
The roots of this (using calculator linked at bottom) are $$s = -1.46,-2.65,-6.87$$
With $1.46$ falling within the root locus as the breakaway point.
However upon emailing my tutor I was advised that the breakaway point was in fact $-1.6$, could someone confirm who is correct? And if my tutor then why?
Calculations for the derivative is here:
Calculations for the roots is here:
https://www.symbolab.com/solver/roots-calculator/roots%202x%5E%7B3%7D%2B18x%5E%7B2%7D%2B48x%2B38
Best Answer
Your approach seems correct. Namely, near a breakaway point two real poles should "meetup" and become a pair of complex conjugate poles or vice versa, thus at a breakaway point one would have repeated (real) poles. Repeated poles also means that the derivative of the characteristic equation with respect to $s$ at the value of the pole should be equal to zero. For example define the general openloop as
$$ H_{ol}(s) = k \frac{N(s)}{D(s)}, \tag{1} $$
with $N(s)$ and $D(s)$ polynomials in $s$. Thus the characteristic equation would be
$$ D(s) + k\,N(s) = 0. \tag{2} $$
The derivative of the characteristic equation would be
$$ D'(s) + k\,N'(s) = 0, \tag{3} $$
with $X'(s)$ denoting the derivative of $X(s)$ with respect to $s$. However, inorder to assure that $s$ is a pole of $(2)$ one can indeed use
$$ k = -\frac{D(s)}{N(s)}. \tag{4} $$
It can be noted that the $k$ in $(2)$ should not change value and thus do not contribute additional terms to the derivative in $(3)$. Substituting $(4)$ into $(3)$ yields
$$ D'(s) - \frac{D(s)}{N(s)}N'(s) = \frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)} = 0. \tag{5} $$
Since $N(s)$ is a polynomial $(5)$ is equivalent to
$$ D'(s)\,N(s) - D(s)\,N'(s) = 0. \tag{6} $$
Even though it is stated that $k$ should remain constant, it can be shown that $(6)$ is equivalent to setting the derivative of $k$ with respect to $s$ to zero. Namely, by using the quotient rule it can be shown that the derivative of $k$ is equal to
$$ k' = -\frac{D'(s)\,N(s) - D(s)\,N'(s)}{N(s)^2}. \tag{7} $$
When using again that $N(s)$ is a polynomial yields that setting $(7)$ to zero is equivalent to $(6)$.
Substituting your openloop transfer function does indeed yield
$$ 2s^3+18s^2+48s+38=0. \tag{8} $$
The only thing I could remark on is that you incorrectly rounded the corresponding roots of $(8)$. For example the first root rounded to two decimal places should be $s=-1.47$.
It can be noted that the other roots of $(8)$ are also breakaway points. Only, if you substitute those values into $(4)$ yields negative values for $k$, while normally a rootlocus plot considers only $k\in [0,\infty)$.