# Proving a function is continuous for all real numbers

This a homework question:

Prove $f(x) = 2x^3 - 4x^2 + 5$ is continuous for all real numbers.

Which proof technique do I use for this? I don't really know where to start.

Hint: If $f$ and $g$ are continuous, then so are $f + g, f - g, f \cdot g, f / g$ (the later provided $g$ isn't equal to $0$).

Every polynomial is locally Lipschitz. More explicitly, take your example $f(x) = 2x^3 - 4x^2 + 5$. Then $$|f(x)-f(x_0)| = |(x-x_0) 2(x^2+x_0 x+x_0^2) - (x-x_0)4(x+x_0)| \le |x-x_0| L$$ for a suitable value of $L$ obtained by using $|x|<|x_0|+\delta$ and the triangle inequality. You can then take $\delta=\varepsilon/L$.

In general, use that $x^n-x_0^n= (x-x_0)(x^{n-1}+x_0 x^{n-2} + \cdots + x_0^{n-2}x+x_0^{n-1})$ to factor $x-x_0$ out of $f(x)-f(x_0)$.

• Polynomials are usually not Lipschitz (on unbounded domains)! However, they are locally Lipschitz... – Dirk Dec 5 '11 at 13:09
• @Dirk, fixed, thanks. – lhf Dec 5 '11 at 13:23

If you know how to prove that the identity function $f(x) = x$ is continuous, then by the algebra of continuous functions you have every polynomial continuous as they are just linear combinations of power functions i.e. $x^n$. If we have $f(x)=x$ continuous, then by the algebra of continuous functions $f\cdot f$ is continuous yielding $x^n$ continuous for any $n\in \mathbb{N}$

To show $f(x) = x$ is continuous consider

$$|f(x_0)- f(x)|=|x_0-x|< \epsilon.$$

Hence, choosing $\delta := \epsilon$ gives the continuity of any polynomial function.

• A polynomial is not a linear combination of the identity function. – lhf Dec 5 '11 at 12:13
• Good point, I should probably change that. I meant to say linear combination of power functions. – Henry Shearman Dec 5 '11 at 13:34

As $f(x)=2x3−4x2+5$ is a polynomial function, it is continuous at every real number (it is a theorem).