1
$\begingroup$

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

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

continuous?

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.