Is this function continuous for $a$?

Question

I need to determine $a\in\mathbb{R}$ so that $f:\mathbb{R}\to\mathbb{R}, f(x):=\begin{cases}x^4-x^2+4a & x\leq 1\\x^3+2x+a^2 &x>1\end{cases}$ is continuous in $\mathbb{R}$.

Idea: I only need to make $f$ continuous in $x=1$: $$ \lim_{x\nearrow1}=4a \text{ and } \lim_{x\searrow1}=3+a^2, \text{ so it has to be }4a=3+a^2.$$ Solving $a^2-4a+3=0$ by using the p-q-fomula I get $a=1$ or $a=3$. By checking $f(1)=4=\lim_{x\to1}f(x)$ for $a=1$ and $f(1)=12=\lim_{x\to1}f(x)$ for $a=3$ I can conclude: $f$ is continuous on $\mathbb{R}$ for all $a\in\{1,3\}$, correct?

There is an error here $$lim_{x\nearrow1}=4a \text{ and } \lim_{x\searrow1}=3+a^2, \text{ so it has to be }4a=3+a^2. $$It should be like this $$lim_{x\nearrow1}=3+a^2 \text{ and } \lim_{x\searrow1}=4a, \text{ so it has to be }4a=3+a^2.$$ — Moustapha_M_I, Apr 15 '23 at 12:48
@Moustapha_M_I ah yes thank you! I mixed up left and right limit... — user1049882, Apr 15 '23 at 12:54

score 1 · Accepted Answer · answered Apr 15 '23 at 12:52

1

What you say is correct.

You only need to check continuity at 1 as elsewhere, the function f is a polynomial and is continuous.

At x = 1, the left hand limit, using the definition $x^4−x^2+4a$ is 1 - 1 + $4a$ = $4a$ and the right hand limit using the definition $x^3+2x+a^2$ is 1 + 2 + $a^2$ = $3 + a^2$.

f is continuous at 1 if the left and right limits at 1 exist and are equal. Solving for the root of the quadratic $a^2 + 3 - 4a = 0$, we get $a = 1$ or $a = 3$. f is continuous on $\mathbb{R}$ if $a = 1$ or $a = 3$ .

answered Apr 15 '23 at 12:52

Mewbacca

51

2

You actually should also check that $f(1)$ equals the two limits, but that is true here. – ronno Apr 15 '23 at 13:08
You are right, however that is covered by the first piecewise definition. – Mewbacca Apr 15 '23 at 14:09

Is this function continuous for $a$?

1 Answers1