\(\displaystyle f(x) \ = \ x-\frac{4}{x^{2}}, \ f'(x) \ = \ 1+\frac{8}{x^{3}} \ = \ 0 \ \implies \ x \ = \ -2\)
\(\displaystyle f(-2) \ = \ -3, \ f'(-2) \ = \ 0 \ = \ m, \ hence \ y \ = \ -3 \ (horizontal \ tangent), \ see \ graph.\)
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