I know there is a similar question. I had a read through it and it didn't help me so I'm posting this one.

The question is

for what value(s) of the constant $a\in \mathbb R$ is

$$f_a(x) = \left\{ \matrix{ax + 1 & \text{if } (x \leqslant 1) \\ (ax)^2 -1 & \text{otherwise}} \right.$$


I've figured out that $a = -1$ or $2$ but have had no luck when subbing those values into the original equation to see which one is continuous.

Could someone tell me if I'm right and what I should do.


Both the functions $ax + 1$ and $(ax)^2 - 1$ are polynomials. Individually, they are continuous for all $x$ no matter which $a$ you pick. The problem is making sure $ax + 1 = (ax)^2 - 1$ when $x = 1$. That is, you wish to solve $a + 1 = a^2 - 1$, which has solutions $a = -1$ and $a = 2$, as you found. These values make your function continuous and are the only such values.


Okay, so we have the function:

$ f(x) = \begin{cases} \hfill ax + 1 \hfill & x \leq 1 \\ \hfill (ax)^2 -1 \hfill & x > 1 \\ \end{cases} $

Both the components of this piecewise function are continuous for all $x$, so the only spot where there might be a discontinuity is where they join, namely at $x = 1$. The task then is to choose $a$ so that the value of both functions at $x=1$ is the same. Therefore:

$a(1) + 1 = (a(1))^2 - 1$

$a + 1 = a^2 - 1$

You correctly deduce that $a = -1$ and $a = 2$ are solutions to this equation. Both make the function continuous, so either is acceptable.

  • $\begingroup$ That should be $a(1)+1=(a\cdot 1)^2-1=a^2-1$ unless the Q was re-written. Some people think that if your answer was posted after the most recent edit of it that it must be your mistake. I might read a Q, think about it, feed the cats, answer the phone, type an answer, check the hockey scores,spend 15 minutes re-writing and removing a horde of typos, and post, only to find the Q changed. $\endgroup$ – DanielWainfleet Apr 7 '16 at 22:38
  • $\begingroup$ My first answer was to the original question which then changed to include a in the squared term. I've fixed my answer to reflect the edited question, sorry if there was any confusion. $\endgroup$ – CRice Apr 7 '16 at 23:19

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