This question is a step involved in solving the below question:
Find an open interval about c on which the inequality $|f(x)-L|<\epsilon$ holds.
$f(x) = 4+x-3x^2$, $L=2$, $c=1$, $\epsilon=0.01$
I started on solving the inequality as below:
$\Rightarrow |f(x)-L|<\epsilon$
$\Rightarrow |4+x-3x^2-2|<0.01$
$\Rightarrow |2+x-3x^2|<0.01$
$\Rightarrow -0.01<2+x-3x^2<0.01$
$\Rightarrow -0.01<2+x-3x^2$ and $2+x-3x^2 < 0.01$
To use the Wavy curve method, I have to make one of the sides of these inequalities zero:
$\Rightarrow 0<2.01+x-3x^2$ and $1.99+x-3x^2 < 0$
However, I am unable to factorize it. How do I solve this inequality so that I can get the answer as an interval of $x$?
Start by looking at the function $-3x^2+x+2$. This is an upside down parabola, with roots $-2/3$ and $1$. The maximum is at $x=1/6$ and has a value of $25/12$. So you will get two intervals, next to each root. One end of the intervals is given by $-3x^2+x+2=0.01$ (these will be the inner limits), the other is given by $-3x^2+x+2=-0.01$ (outer limits). Use the quadratic formula to solve these equations. In the image below I use $\pm0.1$ instead of $\pm0.01$, for clarity