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

The extreme f(x)=(2x^2)/(x-2),(-2,1) is Maximum(0,0)

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

extreme points

f (x) = \frac{2 x ^{2}}{x - 2}, (- 2, 1)

Maximum (0, 0)

Solution steps

Find the critical points:

x = 0

Domain of

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = 0

into

\frac{2 x ^{2}}{x - 2} \Rightarrow y = 0

Maximum (0, 0)

f (x) = 250 (0.9)^{x}, 0 \leq x \leq 6

y ln (x) + x y^{2}

2 x^{3} - 9 x^{2} + 12 x - 8

f (x) = 4 x^{2} - x y + 2 y^{2}

f (x) = \frac{- 3 x + y - 4}{4 x ^{2} - 3 y + 2}

What is the extreme f(x)=(2x^2)/(x-2),(-2,1) ?
The extreme f(x)=(2x^2)/(x-2),(-2,1) is Maximum(0,0)