What is the extreme f(x)=(8x^2)/(x+1),(-1,5) ?

The extreme f(x)=(8x^2)/(x+1),(-1,5) is Minimum(0,0)

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extreme f(x)=(8x^2)/(x+1),(-1,5)

extreme

Minimum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

\frac{8 x ^{2}}{x + 1}, (- 1, 5) : - 1 < x < 5

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 5

Plug the extreme point

x = 0

into

\frac{8 x ^{2}}{x + 1} \Rightarrow y = 0

Minimum (0, 0)

What is the extreme f(x)=(8x^2)/(x+1),(-1,5) ?
The extreme f(x)=(8x^2)/(x+1),(-1,5) is Minimum(0,0)