Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice: 
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$. 
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sure where to go from here. What I tried was the following:
$(1/3)|x+3| < 2$, so $|x+3|<6$
Since the maximum of $x$ is $7/3$, the most $|x+3|$ can be is $|(7/3)+3|= 16/3 < 6$.
Then, the minimum of $x$ is $5/3$, so the least $|x+3|$ can be is $|(5/3)+3| = 14/3 < 6$.
Thus, $|x-2||x+3| < 2$.
Anyone know if this is correct? If not, can you point me in the right direction? Thanks in advance.
 A: Taking from your work. What remains to be shown is: $|x+3| < 6$. By the triangle inequality:
$|x+3| = |x-2+5| \le |x-2|+ |5| < \dfrac{1}{3}+5 < 6$. 
A: Another possible decomposition can be achieved by completing the square and using the triangle inequality: $$|x^2+x-6|=|(x-2)^2+5x-10|\le |x-2|^2+5|x-2|\le \frac 19+\frac 53<2$$
A: By the given $$\frac{5}{3}<x<\frac{7}{3}$$ and we need to prove that:
$$4<x^2+x<8,$$ which is true because $x^2+x$ increases for $x>0$:
$$x^2+x<\left(\frac{7}{3}\right)^2+\frac{7}{3}=\frac{70}{9}<8$$ and $$x^2+x>\left(\frac{5}{3}\right)^2+\frac{5}{3}=\frac{40}{9}>4.$$
A: Let $x \in \mathbb R$ and suppose that $|x−2| < \dfrac 13$. Prove that $|x^2+x−6| < 2$.
\begin{align}
   |x-2| < \dfrac 13 
    &\implies \dfrac 53 < x < \dfrac 73 \\
    &\implies 3+\dfrac 53 < x+3 < 3 + \dfrac 73 \\
    &\implies -\dfrac{16}{3} < x+3 < \dfrac{16}{3} \\
    &\implies |x+3| < \dfrac{16}{3} \\
\end{align}
Hence $|x^2 + x - 6| = |x-2||x+3| < \dfrac 13 \cdot \dfrac{16}{3} < \dfrac{16}{9} < 2$.
