What is the extreme (-8x^3+5x^2-1)/(2x^2-9x) ?

The extreme (-8x^3+5x^2-1)/(2x^2-9x) is Minimum(0.52015…,0.18671…),Maximum(8.67837…,-66.92038…)

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extreme (-8x^3+5x^2-1)/(2x^2-9x)

extreme points

\frac{- 8 x ^{3} + 5 x ^{2} - 1}{2 x ^{2} - 9 x}

Minimum (0.52015 \dots, 0.18671 \dots), Maximum (8.67837 \dots, - 66.92038 \dots)

Solution steps

Find the critical points:

x \approx 0.52015 \dots, x \approx 8.67837 \dots

Domain of

\frac{- 8 x ^{3} + 5 x ^{2} - 1}{2 x ^{2} - 9 x} : x < 0 or 0 < x < \frac{9}{2} or x > \frac{9}{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 0.52015 \dots,

Increasing

: 0.52015 \dots < x < \frac{9}{2},

Increasing

: \frac{9}{2} < x < 8.67837 \dots,

Decreasing

: 8.67837 \dots < x < \infty

Plug the extreme point

x = 0.52015 \dots

into

\frac{- 8 x ^{3} + 5 x ^{2} - 1}{2 x ^{2} - 9 x} \Rightarrow y = 0.18671 \dots

Minimum (0.52015 \dots, 0.18671 \dots)

Plug the extreme point

x = 8.67837 \dots

into

\frac{- 8 x ^{3} + 5 x ^{2} - 1}{2 x ^{2} - 9 x} \Rightarrow y = - 66.92038 \dots

Maximum (8.67837 \dots, - 66.92038 \dots)

Minimum (0.52015 \dots, 0.18671 \dots), Maximum (8.67837 \dots, - 66.92038 \dots)

f (x) = x - 6 x + 9

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

x^{2} - 10 x - 9, 2 \leq x \leq 7

0.002 x^{2} + 4.4 x - 90

f (x) = 7 - 4 x^{2}

What is the extreme (-8x^3+5x^2-1)/(2x^2-9x) ?
The extreme (-8x^3+5x^2-1)/(2x^2-9x) is Minimum(0.52015…,0.18671…),Maximum(8.67837…,-66.92038…)