What is the extreme f(x)=(3x^2-16x+16)/((x^2-2x)^2) ?

The extreme f(x)=(3x^2-16x+16)/((x^2-2x)^2) is Maximum(5.40482…,0.05066…)

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extreme f(x)=(3x^2-16x+16)/((x^2-2x)^2)

extreme

Maximum (5.40482 \dots, 0.05066 \dots)

Solution steps

Find the critical points:

x \approx 5.40482 \dots

Domain of

\frac{3 x ^{2} - 16 x + 16}{( x ^{2} - 2 x ) ^{2}} : x < 0 or 0 < x < 2 or x > 2

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < 5.40482 \dots,

Decreasing

: 5.40482 \dots < x < \infty

Plug the extreme point

x = 5.40482 \dots

into

\frac{3 x ^{2} - 16 x + 16}{( x ^{2} - 2 x ) ^{2}} \Rightarrow y = 0.05066 \dots

Maximum (5.40482 \dots, 0.05066 \dots)

What is the extreme f(x)=(3x^2-16x+16)/((x^2-2x)^2) ?
The extreme f(x)=(3x^2-16x+16)/((x^2-2x)^2) is Maximum(5.40482…,0.05066…)