Prove that the function is continuous

Question

Let $f$ be a given function such that $(f(x))^3 + 2f(x) = x + 1$ for every real $x$. Prove that $f$ is continuous on $\mathbb{R}$.

(I have been trying to prove this, but I find it difficult proving that $f$ is continuous, if we know that the inverse function is.)

Probably better to write $(f(x))^3$. The convention $\sin^3 x = (\sin x)^3$ is a specific convention for trigonometric functions that does not always carry over to other functions. In another context, $f^3(x)$ might mean $f(f(f(x)))$. — David K, Oct 26 '16 at 14:20

score 2 · Answer 1 · answered Oct 26 '16 at 13:39

Verify that:

$$(f(x)-f(y))(f^2(x) + f(x)f(y) + f^2(y) + 2) = x - y$$

Note that $f^2(x) + f(x)f(y) + f^2(y) = \left(f(x) +\frac{f(y)}2\right)^2 + \frac34 f^2(y) \ge0$. We get

$$|f(x)-f(y)| = \frac{|x-y|}{f^2(x) + f(x)f(y) + f^2(y)+2} \le \frac12 |x-y|$$

So $f$ is Lipschtiz continuous.

Mikhail Katz · Answer 2 · 2016-10-26T13:45:06.933

0

$y^3+2y=x+1$ therefore $\frac{dx}{dy}=3y^2+2>0.$ Thus $x$ is a strictly increasing differentiable function of $y$ with positive derivative. By chain rule $\frac{dy}{dx}=\frac{1}{3y^2+2}\leq\frac{1}{2}.$ Therefore $y$ is a differentiable and $\frac{1}{2}$-Lipschitz continuous function of $x$ (there is no need to prove continuity of the inverse function).

edited Oct 26 '16 at 13:45

answered Oct 26 '16 at 13:22

Mikhail Katz

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Prove that the function is continuous

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