Question

Prove (using the "epsilon - delta" definition) that the function f(x) = 4x^2 - 5x + 3 is continuous at every point X_0.

Answer 1

Conditions

Prove (using the "epsilon - delta" definition) that the function $f(x) = 4x^2 - 5x + 3$ is continuous at every point $x_0$.

Solution

Consider function:

$$f(x), x \in [\alpha, b]$$

This function is continuous in $x_0 \in [\alpha, b]$, if:

$$\forall \varepsilon > 0 \ \exists \delta = \delta(\varepsilon) > 0 \ \forall x: |x - x_0| < \delta \ |f(x) - f(x_0)| < \varepsilon$$

Fix $x_0 \in [\alpha, b]$, fix $\varepsilon > 0$

Consider $|f(x) - f(x_0)|$:

$$\begin{array}{l} |f(x) - f(x_0)| = |4x^2 - 5x + 3 - 4x_0^2 + 5x_0 - 3| = |4x^2 - 5x - 4x_0^2 + 5x_0| = \\ = |4x^2 - 5x - 4x_0^2 + 5x_0| \leq 4|x - x_0|^2 + 5|x - x_0| \leq 8\delta|b| + 5\delta < \varepsilon \\ \delta < \frac{\varepsilon}{8|b| + 5} \end{array}$$