# Continuous function

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• Aug 27th 2011, 11:31 PM
terrorsquid
Continuous function
So I'm trying to solve for a and b so that the following function is continuous:

http://i1097.photobucket.com/albums/...continuous.jpg

I was a little confused as to how the $x \in 0$ part applies. Here's what I have done:

$as~x \rightarrow 0^+$ $fe(x) \rightarrow 2$

$\therefore$ for the function to be continuous at $x = 0$, $a$ must = $(0)^3 + 2$

$=2$

$as~x \rightarrow \left(\frac{1}{\pi}\right)^+$ $(bx^2 +2) \rightarrow b\left(\frac{1}{\pi}\right)^2+2$

$\therefore$ for the function to be continuous at $x = \frac{1}{\pi}$, $b$ must = $\frac{\left(\frac{1}{\pi}\right)^2 cos(\pi)}{\left(\frac{1}{\pi}\right)^2}$

$= cos(\pi)$
• Aug 29th 2011, 06:44 PM
terrorsquid
Re: Continuous function
Have I made the function continuous with those values for a and b or am I missing something? I wasn't familiar with finding values to support an interval as opposed to just one value in between the left and right functions.

Also, am I correct in assuming that $x\in 0 < x \le \frac{1}{\pi}$ is the same as $0 ?
• Aug 29th 2011, 08:30 PM
abhishekkgp
Re: Continuous function
Quote:

Originally Posted by terrorsquid
Have I made the function continuous with those values for a and b or am I missing something? I wasn't familiar with finding values to support an interval as opposed to just one value in between the left and right functions.

Also, am I correct in assuming that $x\in 0 < x \le \frac{1}{\pi}$ is the same as $0 ?

although i have never seen something like $x \in 0 < x <1$ but i am assuming this means $x \in (0,1)$, which is what you have done too. Also, yes, you have made the function continuous in $\mathbb{R}$ by these values of $a$ ad $b$.
• Aug 30th 2011, 08:54 AM
HallsofIvy
Re: Continuous function
Quote:

Originally Posted by terrorsquid
So I'm trying to solve for a and b so that the following function is continuous:

http://i1097.photobucket.com/albums/...continuous.jpg

I was a little confused as to how the $x \in 0$ part applies.

That's because there is NO " $x \in o$". There is, rather, $x\in 0< x\le\frac{1}{\pi}$. That is, x is in the interval from 0 to [itex]\frac{1}{\pi}, including $\frac{1}{\pi}$, but not 0.

Quote:

Here's what I have done:

$as~x \rightarrow 0^+$ $fe(x) \rightarrow 2$

$\therefore$ for the function to be continuous at $x = 0$, $a$ must = $(0)^3 + 2$

$=2$

$as~x \rightarrow \left(\frac{1}{\pi}\right)^+$ $(bx^2 +2) \rightarrow b\left(\frac{1}{\pi}\right)^2+2$

$\therefore$ for the function to be continuous at $x = \frac{1}{\pi}$, $b$ must = $\frac{\left(\frac{1}{\pi}\right)^2 cos(\pi)}{\left(\frac{1}{\pi}\right)^2}$

$= cos(\pi)$