contestada

The function f(x)=x+3x2−9 is defined everywhere except at x=±3. If possible, define f at x=−3 so that it becomes continuous at x=−3.

Respuesta :

sqdancefan

The function simplifies to

[tex]f(x)=\dfrac{x+3}{x^2-9}=\dfrac{x+3}{(x+3)(x-3)}=\dfrac{1}{x-3}\qquad\text{except at $x=-3$}[/tex]

By defining f(-3) to be 1/(-3-3) = -1/6, we can plug the "hole" at that point, making the function continuous there.

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The attachments show the original function being undefined at x=-3. Defining the function at x=-3 removes the discontinuity.

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