# Continuous function

• Aug 27th 2011, 10:31 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 $\displaystyle x \in 0$ part applies. Here's what I have done:

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

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

$\displaystyle =2$

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

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

$\displaystyle = cos(\pi)$
• Aug 29th 2011, 05:44 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 $\displaystyle x\in 0 < x \le \frac{1}{\pi}$ is the same as $\displaystyle 0<x\le \frac{1}{\pi}$ ?
• Aug 29th 2011, 07:30 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 $\displaystyle x\in 0 < x \le \frac{1}{\pi}$ is the same as $\displaystyle 0<x\le \frac{1}{\pi}$ ?

although i have never seen something like $\displaystyle x \in 0 < x <1$ but i am assuming this means $\displaystyle x \in (0,1)$, which is what you have done too. Also, yes, you have made the function continuous in $\displaystyle \mathbb{R}$ by these values of $\displaystyle a$ ad $\displaystyle b$.
• Aug 30th 2011, 07:54 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 $\displaystyle x \in 0$ part applies.

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

Quote:

Here's what I have done:

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

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

$\displaystyle =2$

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

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

$\displaystyle = cos(\pi)$