Problem 890

Define g(7) for the given function so that it is continuous at x = 7.

g(x) = \frac{4x^{2}-196}{4x-28}




The function g(x) = \frac{4x^{2}-196}{4x-28} is undefined when the denominator is 0.

So, the function is undefined at x = 7.

\frac{4x^{2}-196}{4x-28} can be simplified by dividing numerator and denominator by 4, factoring the numerator, and dividing the common factor in the numerator and denominator.

\frac{4x^{2}-196}{4x-28} =  \frac{x^{2}- 49}{x-7}  = \frac{(x+7)(x-7)}{(x-7)} = x+7

The graphs of y = \frac{4x^{2}-196}{4x-28} and y = x+7 are the same expect that the graph of y = \frac{4x^{2}-196}{4x-28} has a hole at x = 7.

This hole can be eliminate by defining g(7) to be the value of x + 7 on the graph of y = x + 7.

Thus , g(7) should be define as 14.




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