# $f(x) =\ln(2x^2 + 1)$ is continuous on $\mathbb{R}$

True or False

The function $f : \Bbb R \to \Bbb R$ deﬁned by $f(x) = \ln(2x^2 + 1)$ is continuous on $\Bbb R$.

I know this condition that

The function $f$ is continuous at some point $c$ of its domain if the limit of $f(x)$ as $x$ approaches $c$ through the domain of $f$ exists and is equal to $f(c)$.

How to find that point c for above question and How to proceed further ?

• In general if two functions are continuous, then their composition is also continuous. Apr 21, 2015 at 4:53
• But how does it is applied here ? I dont think there is any composite function here Apr 21, 2015 at 4:55
• Well, if we say $f$ is continuous on $\Bbb R$, then we mean $f$ is continuous for every $c \in \Bbb R$. So there is no finding just one point $c$. It's true for every $c$. Are you OK with the fact that $h(x) = \ln{x}$ is continuous, and $g(x) = 2x^{2} + 1$ is continuous? If so, the $f(x)$ you wrote is just $h(g(x))$, so it is the composition of two continuous functions. Apr 21, 2015 at 4:55
• @user46944 THen How to proceed for this question Apr 21, 2015 at 4:56
• Think about the two functions $\ln x$ and $2x^2+1$. Apr 21, 2015 at 4:56

Lets express $f(x)$ as $f_1(x)$ and $f_2(x)$, where $f_1(x)=ln(x$) and $f_2(x)=2x^2+1$, and so $f(x)=f_1(f_2(x))$
Ok. It is pretty obvious that $f(x)$ is defined over whole domain of real numbers. $f_1(x)$ is defined in $(0,\infty)$ domain, and it does happen so that $f_2(x)$ range is $[1,\infty)$ which fits into $(0,\infty)$ quite nicely.
Only point of somewhat interest is when $x=0$, which doesn't make any difference because $f_2(x)$ behaves very nicely there, still just to be sure
$\lim_{x\to+0} f_2(x)=1$ and $\lim_{x\to-0} f_2(x)=1$, with $f_2(0)=1$