The local minima of $f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$ are (x, f(x)) = (-1.5, 0) and (7.980, 609.174)

<h3>How to determine the local minima?</h3>

The function is given as:

$f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}$

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

Minimum = (-1.5, 0)

Minimum = (7.980, 609.174)

The above means that, the local minima are

(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

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