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

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

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

\frac{15 - 4}{2} \leq x < 0 or x > 0

Interval Notation

[\frac{15 - 4}{2}, 0) \cup (0, \infty)

Decimal Notation

- 0.06350 \dots \leq x < 0 or x > 0

Solution steps

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

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq \frac{\frac{1}{x + 4} - \frac{1}{4}}{x} and \frac{\frac{1}{x + 4} - \frac{1}{4}}{x} \leq x

- 2 \leq \frac{\frac{1}{x + 4} - \frac{1}{4}}{x} : x < - 4 or - \frac{31}{8} \leq x < 0 or x > 0

\frac{\frac{1}{x + 4} - \frac{1}{4}}{x} \leq x : - 4 < x \leq - \frac{4 + 15}{2} or x \geq \frac{15 - 4}{2}

Combine the intervals

(x < - 4 or - \frac{31}{8} \leq x < 0 or x > 0) and (- 4 < x \leq - \frac{4 + 15}{2} or x \geq \frac{15 - 4}{2})

Merge Overlapping Intervals

\frac{15 - 4}{2} \leq x < 0 or x > 0

m^{4} + m^{3} + m^{2} + m

5 ab + 10 a c + 15 a

\frac{x - 4}{x ^{2} - 36} < 0

(- 3 x)^{\frac{1}{3}}

x^{2} = \frac{1}{2}

What is -2<= (1/(x+4)-1/4)/x <= x ?
The solution to -2<= (1/(x+4)-1/4)/x <= x is (sqrt(15)-4)/2 <= x<0\lor x>0