What is the extreme f(x)=(4x+3)/(x^2) ?

The extreme f(x)=(4x+3)/(x^2) is Minimum(-3/2 ,-4/3)

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extreme f(x)=(4x+3)/(x^2)

extreme

Minimum (- \frac{3}{2}, - \frac{4}{3})

Solution steps

f^{'} (x) = \frac{2 ( - 2 x - 3 )}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - \frac{3}{2}

into

\frac{4 x + 3}{x ^{2}} : - \frac{4}{3}

Minimum (- \frac{3}{2}, - \frac{4}{3})

What is the extreme f(x)=(4x+3)/(x^2) ?
The extreme f(x)=(4x+3)/(x^2) is Minimum(-3/2 ,-4/3)