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Mathematics 16 Online
OpenStudy (anonymous):

give a formula for the extended function that is continuous at the indicated point: f(x) = ( x^2 - 9 ) / ( x + 3 ) , x = -3

OpenStudy (anonymous):

Use a piecewise function to specify when x = 3.

OpenStudy (anonymous):

hmm. well i know that the answer is y = x - 3, i just have no idea how to get there.

OpenStudy (anonymous):

You'd need to add the point at \(x=-3\), using the definition of continuity. So we have: \[ \lim_{x\to -3}\frac{x^2-9}{x+3}=f(-3) \]With this, we solve: \[ \lim_{x\to -3}\frac{x^2-9}{x+3}=\lim_{x\to -3}\frac{(x+3)(x-3)}{x+3}=\lim_{x\to -3}x-3=-6 \]Therefore, at \(x=-3, f(x)=-6\), so we simply add the point to the function to get: \[ f(x)=\begin{cases} \frac{x^2-9}{x+3}, x\ne -3\\\\ x=-6, x=-3 \end{cases} \]

OpenStudy (anonymous):

x2−9x+3,x≠−3 x=−6,x=−3 x=−6,x=−3 x=−6,x=−3 x=−6,x=−3

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