What is the extreme f(x)= 8/(x^2)+1/((40-x)^2) ?

The extreme f(x)= 8/(x^2)+1/((40-x)^2) is Minimum(80/3 , 27/1600)

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extreme f(x)= 8/(x^2)+1/((40-x)^2)

extreme

Minimum (\frac{80}{3}, \frac{27}{1600})

Solution steps

f^{'} (x) = - \frac{16}{x ^{3}} + \frac{2}{( 40 - x ) ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{80}{3},

Increasing

: \frac{80}{3} < x < 40,

Decreasing

: 40 < x < \infty

Plug

x = \frac{80}{3}

into

\frac{8}{x ^{2}} + \frac{1}{( 40 - x ) ^{2}} : \frac{27}{1600}

Minimum (\frac{80}{3}, \frac{27}{1600})

What is the extreme f(x)= 8/(x^2)+1/((40-x)^2) ?
The extreme f(x)= 8/(x^2)+1/((40-x)^2) is Minimum(80/3 , 27/1600)