What is (12(3x^2-3x^6))/((1+3x^4)^2)>0 ?

The solution to (12(3x^2-3x^6))/((1+3x^4)^2)>0 is -1<x<0\lor 0<x<1

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(12(3x^2-3x^6))/((1+3x^4)^2)>0

\frac{12 ( 3 x ^{2} - 3 x ^{6} )}{( 1 + 3 x ^{4} ) ^{2}} > 0

- 1 < x < 0 or 0 < x < 1

Interval Notation

(- 1, 0) \cup (0, 1)

Solution steps

\frac{12 ( 3 x ^{2} - 3 x ^{6} )}{( 1 + 3 x ^{4} ) ^{2}} > 0

Rewrite in standard form

\frac{3 x ^{2} - 3 x ^{6}}{( 1 + 3 x ^{4} ) ^{2}} > 0

Factor

Multiply both sides by

- 1

(reverse the inequality)

Simplify

Divide both sides by

3

Simplify

Identify the intervals

- 1 < x < 0 or 0 < x < 1

\frac{2 x + 3}{5} < 10

- 12 = - 3 + \frac{x}{2}

\frac{3 ^{\frac{3}{5}}}{3 ^{\frac{2}{5}}}

- 6 - (- 6)

x - 1^{2}

What is (12(3x^2-3x^6))/((1+3x^4)^2)>0 ?
The solution to (12(3x^2-3x^6))/((1+3x^4)^2)>0 is -1<x<0\lor 0<x<1