What is the critical (x^3+6x-8)/x-3x ?

The critical (x^3+6x-8)/x-3x is x\approx-1.21401…

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critical (x^3+6x-8)/x-3x

critical points

\frac{x ^{3} + 6 x - 8}{x} - 3 x

x \approx - 1.21401 \dots

Solution steps

\frac{x ^{3} + 6 x - 8}{x} - 3 x

Find where

f^{'} (x)

is equal to zero or undefined

x \approx - 1.21401 \dots, x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{3} + 6 x - 8}{x} - 3 x : x < 0 or x > 0

\frac{x ^{3} + 6 x - 8}{x} - 3 x

is not defined at

x = 0

, therefore:

x \approx - 1.21401 \dots

\frac{- 3 x ^{2} + 24 x - 45}{2 x ^{2} - 10 x}

4 x - 3 y = 6

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

(x - 2)^{2} - 1

f (x) = \frac{3 x - 4}{5 x + 3}

What is the critical (x^3+6x-8)/x-3x ?
The critical (x^3+6x-8)/x-3x is x\approx-1.21401…