What is (x^2)/(1+x^4)<= 1/2 ?

The solution to (x^2)/(1+x^4)<= 1/2 is True for all x

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(x^2)/(1+x^4)<= 1/2

\frac{x ^{2}}{1 + x ^{4}} \leq \frac{1}{2}

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

\frac{x ^{2}}{1 + x ^{4}} \leq \frac{1}{2}

Rewrite in standard form

\frac{2 x ^{2} - ( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )} \leq 0

Factor

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

\frac{- ( x + 1 ) ^{2} ( x - 1 ) ^{2}}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 1 ) ^{2} ( x - 1 ) ^{2} ) ( - 1 )}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )} \geq 0 \cdot (- 1)

Simplify

\frac{( x + 1 ) ^{2} ( x - 1 ) ^{2}}{( x ^{2} + 2 x + 1 ) ( x ^{2} - 2 x + 1 )} \geq 0

Identify the intervals

True for all x

(- 3 + i)^{4}

\frac{5}{4} w = 40

6 x^{2} + 12 x + 5 = 0

9 x^{2} - 36 x - 45

\frac{y ^{7}}{y ^{3}}

What is (x^2)/(1+x^4)<= 1/2 ?
The solution to (x^2)/(1+x^4)<= 1/2 is True for all x