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Hi, I am going to change the function slightly as I think you have copied it incorrectly. I want the expression to be
The value of -2 is important since the expression \frac{x^2 + 3x + 2}{x + 2} doesn't exist at x = -2 since the denominator is zero. Looking at the expression \frac{x^2 + 3x + 2}{x + 2} more closely you can see that the numerator factors so that \frac{x^2 + 3x + 2}{x + 2} = \frac{(x + 2)(x + 1)}{x + 2}. Thus using cancellation it seems that the expressions \large \frac{x^2 + 3x + 2}{x + 2} and x + 1 are equal but that is not true since the first expression doesn't exist for x = -2 and the second expression has value -1 for x = -2. Now back to the problem of the domain and graph of f(x) = \left( \begin{array}{ll}\frac{x^2 + 3x + 2}{x + 2} & \mbox{if $x \neq -2$}\\4 & \mbox{if $x = -2$}. \end{array} \right. If x \neq -2 then f(x) is evaluated using \large \frac{x^2 + 3x + 2}{x + 2} and for x = -2, \; f(x) = f(-2) = 4 and thus f(x) exists for all real numbers x and thus the domain of f(x) is all real numbers. To graph y = f(x) I would start by graphing y = x + 1 which is the line with slope 1 and passing through (0, 1). y = x + 1 We know that the functions y = x + 1 and y = f(x) have the same values for all x except x = -2. At x = -2 we have f(x) = f(-2) = 4 and thus my graph of $y = f(x) is y = f(x) The small circle on the line at x = -2 is to indicate that there is a "hole" in the line there and the value of f(-2) is 4. I hope this helps, | |||||||||||||||
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