What is the extreme f(x)=(9(t-2)^2)/(t^2) ?

The extreme f(x)=(9(t-2)^2)/(t^2) is Minimum(2,0)

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extreme f(x)=(9(t-2)^2)/(t^2)

extreme

Minimum (2, 0)

Solution steps

f^{'} (t) = \frac{36 ( t - 2 )}{t ^{3}}

Find intervals:

Increasing

: - \infty < t < 0,

Decreasing

: 0 < t < 2,

Increasing

: 2 < t < \infty

Plug

t = 2

into

\frac{9 ( t - 2 ) ^{2}}{t ^{2}} : 0

Minimum (2, 0)

What is the extreme f(x)=(9(t-2)^2)/(t^2) ?
The extreme f(x)=(9(t-2)^2)/(t^2) is Minimum(2,0)