Question

Prove from first principles (i.e.an ε − δ proof) that f is continuous at the point x = −3.

Answer 1

Let $f(x)=4x^2-3x+2$

Suppose we wish to show this is continuous using the $\epsilon − δ$ definition. Then we need

to control the error $|f(x) − f(x_0)| \text{ in terms of } |x − x_0|$ . We have

$|f(x) − f(x_0)| =|(4x2 − 3x + 2) − (4x^2_0 − 3x_0 + 2)|$

        $=|4x^2 − 4x^2_0 − 3x + 3x_0 + 2 − 2| = |4(x − x0)(x + x0) − 3(x − x0)|\\ = |(x − x_0) (4(x + x_0) − 3)|\\ = |x − x_0| |4(x + x_0) − 3|\\≤ |x − x_0|(4|x + x_0| + 3)\\$

     

Thus we have bounded the error$|f(x) − f(x_0)| \text{ in terms of } |x − x_0|$ . We now need to

show that given any $\epsilon> 0$ , we can choose δ > 0 small enough so that if $|x − x_0| < δ,$

then $|f(x) − f(x_0)| < \epsilon.$ Indeed, suppose $|x − x_0| < 1$ . Then we have

$|f(x) − f(x_0)| ≤ |x − x_0|(4|x − x_0| + 8|x_0| + 3) ≤ |x − x_0|(8|x_0| + 7)$

If we further have $|x − x_0| <\epsilon 8|x_0|+7$ , then

$|f(x) − f(x_0)| ≤ |x − x_0|(8|x_0| + 7)$

$\text{at }x_0=3, |f(x) − f(-3)| ≤ |x +3|(8|-3| + 7)\\ |f(x) − f(-3)| ≤ |x +3|(31)$

Hence The function f is continuous.