What is the range of (4x^3-3x^2+2+x)/(x(2x^2-3x+1)) ?

The range of (4x^3-3x^2+2+x)/(x(2x^2-3x+1)) is f(x) 2

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range of (4x^3-3x^2+2+x)/(x(2x^2-3x+1))

range

\frac{4 x ^{3} - 3 x ^{2} + 2 + x}{x ( 2 x ^{2} - 3 x + 1 )}

f (x) \leq - 29.32416 \dots or 0 < f (x) < 2 or f (x) > 2

Interval Notation

(- \infty, - 29.32416 \dots] \cup (0, 2) \cup (2, \infty)

Solution steps

Find the minimum and maximum value in each defined interval and unite the results

Domain of

\frac{4 x ^{3} - 3 x ^{2} + 2 + x}{x ( 2 x ^{2} - 3 x + 1 )} : x < 0 or 0 < x < \frac{1}{2} or \frac{1}{2} < x < 1 or x > 1

Extreme Points of

\frac{4 x ^{3} - 3 x ^{2} + 2 + x}{x ( 2 x ^{2} - 3 x + 1 )} :

Minimum

(0.20778 \dots, 21.97581 \dots),

Maximum

(0.75412 \dots, - 29.32416 \dots)

Find the range for the interval

- \infty < x < 0 : 0 < f (x) < 2

Find the range for the interval

0 < x < \frac{1}{2} : 21.97581 \dots \leq f (x) < \infty

Find the range for the interval

\frac{1}{2} < x < 1 : - \infty < f (x) \leq - 29.32416 \dots

Find the range for the interval

1 < x < \infty : 2 < f (x) < \infty

Combine the ranges of all domain intervals to obtain the function range

0 < f (x) < 2 or f (x) \geq 21.97581 \dots or f (x) \leq - 29.32416 \dots or f (x) > 2

f (x) \leq - 29.32416 \dots or 0 < f (x) < 2 or f (x) > 2

\frac{( x - 6 )}{x + 6}

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

f (x) = x^{3} + 3 x^{2} - 16 x - 48

3 - 2 x - x^{3}

f (x) = 3 x^{2} + 4 y = 12

What is the range of (4x^3-3x^2+2+x)/(x(2x^2-3x+1)) ?
The range of (4x^3-3x^2+2+x)/(x(2x^2-3x+1)) is f(x)<=-29.32416…\lor 0<f(x)<2\lor f(x)>2