Answer
$(-∞,-2)∪ (3, ∞)$
Work Step by Step
Consider the Rational Inequality as follows:
$\frac{1}{x+3}<1$
Here are the steps required for Solving Rational Inequalities:
Step 1: One side must be zero and the other side can have only one fraction, so we simplify the fractions if there is more than one fraction. Let us subtract 1 from both sides to obtain zero on the right.
$\frac{1}{x+3}<1$
$\frac{1}{x+3}-1<0$
$\frac{1-1(x-3)}{x-3}<0$
$\frac{-x-2}{x-3}<0$
Step 2: Critical or Key Values are first evaluated. In order to this, set the numerator and denominator of the fraction equal to zero and then simplified rational inequality is solved.
$-x-2 = 0$
This implies
$x =-2$
And
$x-3=0$
This implies
$ x = 3$
These solutions are used as boundary points on a number line.
Step 3: Locate the boundary points on a number line found in Step 2 to divide the number line into intervals.
The boundary points divide the number line into three intervals:
$(-∞,-2), (-2, 3), (3, ∞)$
Step 4: Now, one test value within each interval is chosen and $f$ is evaluated at that number.
Intervals: $(-∞,-2), (-2, 3), (3, ∞)$
Test value: $-3$ $0$ $4$
Sign Change: Negative Positive Negative
$f (x)< 0?$: T F T
Step 5: Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x)< 0$.
Based on our work done in Step 4, we see that $f (x)< 0$ for all $x$ in
$(-∞,-2) $or $(3, ∞)$ .
Conclusion: Thus, the interval notation of the given inequality is $(-∞,-2)∪ (3, ∞)$ and the graph of the solution set on a number line is shown as follows:
