2
$\begingroup$

I have the following graph:

$h(t) = \begin{cases} 2t+1, & \mathrm{if}\ t \le -1, \\ 3t, & \mathrm{if}\ -1 < t < 1, \\ 2t-1, & \mathrm{if}\ t \ge 1.\end{cases}$

The question I have to answer is: Give the conditions which would make the function $h$ continuous at the point $t = a$.

This question does not make sense to me. Firstly there is no a in the equation, and secondly it does not matter what value i would give to $t$, this function can never be continuous.

Do I misunderstand what is required here?

  • 1
    I wonder why a question should be written in such a complicated language. If the question is "At which points is $h$ continuous?", why is it not formulated like that?2012-08-09

3 Answers 3

2

The question asks for the conditions (on $a$) which make the function continuous at $t=a$. I think you know the answer.

The function is not continuous at $t=-1$. (Perhaps in your answer you should explain how you know this.) It is not continuous at $t=1$. It is continuous at $t=a$ for any value of $a$ other than $-1$ or $1$. (Perhaps you should explain how you know that.)

  • 0
    @MichaelFrey: Note by the way that if we replace the $2t+1$ at the beginning by $2t-1$, and the $2t-1$ at the end by $2t+1$, then the function is continuous everywhere.2012-08-09
1

The only true variable here is $a$ ($t$ is just a dummy variable of function $h$, and $h(t)$ is fixed in definition). Therefore, any restrictions to make $h(t)$ continuous at $t=a$ would be on $a$, not $t$. Plus, the question asks to make $h$ continuous on a single point $a$, not to make the whole function continuous.

Thus, the answer of the question is that $a \neq 1, -1$.

0

so conditions would be that limit of left and right side of function would be same,also $h(a)$ exist and is equal to limit of left and right side;