Not Halfway, But Here

A training job is split across $n$ parallel workers. Adding workers divides the actual computation, but every extra worker also adds coordination overhead, so the model of total wall-clock time is

$T(n) = \frac{180}{n} + 4n \quad \text{minutes}$

Treat $n$ as a continuous variable so the curve is smooth.

Going from $n = 2$ workers to $n = 12$ workers, $T$ has some average rate of change, in minutes per worker. Somewhere strictly between $2$ and $12$ there is exactly one worker count $n^{*}$ at which the instantaneous rate of change of $T$ is equal to that average rate — the one place where the tangent line is exactly as steep as the overall trend.

Find $n^{*}$, rounded to two decimal places.