What is 1+(12}{x+1}<= \frac{12)/x ?

The solution to 1+(12}{x+1}<= \frac{12)/x is -4<= x<-1\lor 0<x<= 3

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1+(12}{x+1}<= \frac{12)/x

1 + \frac{12}{x + 1} \leq \frac{12}{x}

- 4 \leq x < - 1 or 0 < x \leq 3

Interval Notation

[- 4, - 1) \cup (0, 3]

Solution steps

1 + \frac{12}{x + 1} \leq \frac{12}{x}

Rewrite in standard form

\frac{x ^{2} + x - 12}{x ( x + 1 )} \leq 0

Factor

\frac{x ^{2} + x - 12}{x ( x + 1 )} : \frac{( x - 3 ) ( x + 4 )}{x ( x + 1 )}

\frac{( x - 3 ) ( x + 4 )}{x ( x + 1 )} \leq 0

Identify the intervals

- 4 \leq x < - 1 or 0 < x \leq 3

(x - 7)^{\frac{7}{2}} - 49 (x - 7)^{\frac{3}{2}}

12 x^{3} + 48 x^{2} + 48 x

\frac{a}{- 8} + 15 > 23

232

What is 1+(12}{x+1}<= \frac{12)/x ?
The solution to 1+(12}{x+1}<= \frac{12)/x is -4<= x<-1\lor 0<x<= 3