What is the range of (x^4)/(x^2+x-6) ?

The range of (x^4)/(x^2+x-6) is f(x) = (3(1833-67sqrt(201)))/(200)

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range of (x^4)/(x^2+x-6)

range

\frac{x ^{4}}{x ^{2} + x - 6}

f (x) \leq 0 or f (x) \geq \frac{3 ( 1833 - 67 201 )}{200}

Solution steps

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

Domain of

\frac{x ^{4}}{x ^{2} + x - 6} : x < - 3 or - 3 < x < 2 or x > 2

Extreme Points of

\frac{x ^{4}}{x ^{2} + x - 6} :

Minimum

(\frac{- 3 - 201}{4}, \frac{3 ( 1833 + 67 201 )}{200}),

Maximum

(0, 0),

Minimum

(\frac{- 3 + 201}{4}, \frac{3 ( 1833 - 67 201 )}{200})

Find the range for the interval

- \infty < x < - 3 : \frac{3 ( 1833 + 67 201 )}{200} \leq f (x) < \infty

Find the range for the interval

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

Find the range for the interval

2 < x < \infty : \frac{3 ( 1833 - 67 201 )}{200} \leq f (x) < \infty

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

f (x) \geq \frac{3 ( 1833 + 67 201 )}{200} or f (x) \leq 0 or f (x) \geq \frac{3 ( 1833 - 67 201 )}{200}

f (x) \leq 0 or f (x) \geq \frac{3 ( 1833 - 67 201 )}{200}

f (x) = \frac{1}{x - 4}

f (x) = 9 - x^{2} - x + 1

f (x) = \frac{1}{3 x - x ^{2}}

\frac{x ^{2}}{x ^{2} - 16}

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

What is the range of (x^4)/(x^2+x-6) ?
The range of (x^4)/(x^2+x-6) is f(x)<= 0\lor f(x)>= (3(1833-67sqrt(201)))/(200)