What is (a^2)/((a^4+1))>= 1/2 ?

The solution to (a^2)/((a^4+1))>= 1/2 is a=-1\lor a=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(a^2)/((a^4+1))>= 1/2

\frac{a ^{2}}{( a ^{4} + 1 )} \geq \frac{1}{2}

a = - 1 or a = 1

Interval Notation

a = - 1 \cup a = 1

Solution steps

\frac{a ^{2}}{a ^{4} + 1} \geq \frac{1}{2}

Rewrite in standard form

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

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

a = - 1, a = 1

a = - 1 or a = 1

x^{3} - 4 x^{2} + 2 < x^{3} - 2 x^{2} + 1

\frac{- 3}{- x + 7} > 0

\frac{1}{( x - 1 )} \leq 2

3 x^{3} \geq 12

\frac{( x ^{2} - 19 x + 88 )}{( x ^{2} - 11 x + 24 )} \leq 0

What is (a^2)/((a^4+1))>= 1/2 ?
The solution to (a^2)/((a^4+1))>= 1/2 is a=-1\lor a=1