How do solve $\frac{3}{x + 1} \leq 3$ $3/(x+1)<=3$ graphically?

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Answer 1 · 2018-05-25T16:04:11

The solution is $x \in (- \infty, - 1) \cup [0, + \infty)$ $x in (-oo, -1) uu[0, +oo)$

Plot the curve

$f (x) = \frac{3}{x + 1} - 3$ $f(x)=3/(x+1)-3$

The function is undefined for $x = - 1$ $x=-1$

graph{3/(x+1)-3 [-18.02, 18.03, -9.01, 9.01]}

Then, look where $f (x) \leq 0$ $f(x)<=0$

The solution is $x \in (- \infty, - 1) \cup [0, + \infty)$ $x in (-oo, -1) uu[0, +oo)$

How do solve 3x+1≤33/(x+1)<=3 graphically?