In precalculus, you may have written things like: ((x^2 +x-6)/(x+3))=x-2. This seems like f(x)=((x^2 +x-6)/(x+3)) and g(x)=x-2 are he same function. Explain why they’re not.

In precalculus, you might encounter expressions like \(\frac{x^2 + x - 6}{x + 3} = x - 2\). While these two expressions appear similar, they do not define the *same function* for all values of \(x\).

### Why They Are Not the Same Function

1. **Domain Differences**: 
   - The function \(f(x) = \frac{x^2 + x - 6}{x + 3}\) has a restriction on the domain because \(x + 3\) appears in the denominator. When \(x = -3\), the denominator becomes zero, which makes \(f(x)\) undefined at \(x = -3\).
   - In contrast, \(g(x) = x - 2\) has no such restriction, as there is no denominator that could cause a division by zero. Thus, \(g(x)\) is defined for all real numbers.

2. **Simplification Process**:
   - The expression \(\frac{x^2 + x - 6}{x + 3}\) can be simplified to \(x - 2\) for values of \(x \neq -3\), because \(x^2 + x - 6\) factors to \((x + 3)(x - 2)\), allowing us to cancel \(x + 3\) from the numerator and denominator

In precalculus, you may have written things like: ((x^2 +x-6)/(x+3))=x-2. This seems like f(x)=((x^2 +x-6)/(x+3)) and g(x)=x-2 are the same function. Explain why they’re not.

Understand why the functions f(x) = (x^2 + x - 6) / (x + 3) and g(x) = x - 2 are not identical, focusing on domain differences and simplification processes. Explore key precalculus concepts on rational functions and domain restrictions, ideal for Grade 11-12 students.

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