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 points

f (x) = \frac{9 ( t - 2 ) ^{2}}{t ^{2}}

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)

x y - 5 x - 5 y - x^{2} - y^{2}

e^{2 x}

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 12 y + 13

f (x) = - 4 x^{4} + 3 x^{2} - 15 x + 5

f (x) = x^{2} + 3 - x

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)