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Math Help - Continuous function

  1. #1
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    Continuous function

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



    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)
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  2. #2
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    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<x\le \frac{1}{\pi} ?
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Re: Continuous function

    Quote Originally Posted by terrorsquid View Post
    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<x\le \frac{1}{\pi} ?
    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.
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  4. #4
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    Re: Continuous function

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



    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.

    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)
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