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Thread: continuity

  1. #1
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    continuity

    5 Consider the function f with rule f(x) {(x^2 - 25 )/ (x-5) if x ≠ 5
    a if x = 5

    If f is continuous on R, find the value of a.
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  2. #2
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    Quote Originally Posted by rachael
    5 Consider the function f with rule f(x) {(x^2 - 25 )/ (x-5) if x ≠ 5
    a if x = 5

    If f is continuous on R, find the value of a.
    I understand you question as,
    $\displaystyle f(x)=\frac{x^2-25}{x-5},x\not =5$ is a real-valued function. It is definitely not countinous on $\displaystyle \mathbb{R}$ because 5 is not in its domain. Thus, you are asking how do define the function at $\displaystyle x=5$ thus it would be countinous? Notice you function is, $\displaystyle \frac{x^2-25}{x-5}=\frac{(x-5)(x+5)}{x-5}$=$\displaystyle x+5$. This a linear function not defined at $\displaystyle x=5$. Thus, we need to redefine it at $\displaystyle x=5$ call that function $\displaystyle g(x)$. We need not do redefine it anywhere else because it is linear and is countinous. By the definition of countinuity we need that,
    $\displaystyle \lim_{x\to 5}g(x)=g(5)$ Since, $\displaystyle f$ and $\displaystyle g$ are function that agree at all but one point (at 5) we have that they have the same limit. Thus, $\displaystyle \lim_{x\to 5}f(x)=g(5)$, but $\displaystyle \lim_{x\to 5}f(x)=\lim_{x\to 5}x+5=10$. Thus, we have $\displaystyle g(5)=10$. Thus, the function,
    $\displaystyle g(x)=x+5$ is the redefined countinous function of $\displaystyle f(x)=\frac{x^2-25}{x-5}$.
    Last edited by ThePerfectHacker; Apr 6th 2006 at 06:47 AM.
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