Choose The Function That Has Domain $x \neq -3$ And Range $y \neq 2$.A. $f(x)=\frac{x+2}{x+3}$B. $f(x)=\frac{2x+1}{x+3}$C. $f(x)=\frac{x-3}{x+2}$

When dealing with functions, it's essential to understand the concept of domain and range. The domain of a function is the set of all possible input values (x-values) that the function can accept, while the range is the set of all possible output values (y-values) that the function can produce. In this article, we will explore the domain and range of three given functions and determine which one has a domain of $x \neq -3$ and a range of $y \neq 2$.

Understanding the Domain and Range

The domain of a function is the set of all possible input values that the function can accept. This means that the function is defined for all values of x in the domain. On the other hand, the range of a function is the set of all possible output values that the function can produce. This means that the function can produce all values of y in the range.

Analyzing Function A: $f(x)=\frac{x+2}{x+3}$

Let's start by analyzing function A: $f(x)=\frac{x+2}{x+3}$. To determine the domain of this function, we need to find the values of x that make the denominator equal to zero. In this case, the denominator is $x+3$, which equals zero when $x=-3$. Therefore, the domain of function A is all real numbers except $x=-3$.

To determine the range of function A, we need to find the values of y that the function can produce. Since the function is a rational function, we can use the concept of horizontal asymptotes to determine the range. The horizontal asymptote of a rational function is the value that the function approaches as x approaches infinity. In this case, the horizontal asymptote is $y=1$. Therefore, the range of function A is all real numbers except $y=2$.

Analyzing Function B: $f(x)=\frac{2x+1}{x+3}$

Now, let's analyze function B: $f(x)=\frac{2x+1}{x+3}$. To determine the domain of this function, we need to find the values of x that make the denominator equal to zero. In this case, the denominator is $x+3$, which equals zero when $x=-3$. Therefore, the domain of function B is all real numbers except $x=-3$.

To determine the range of function B, we need to find the values of y that the function can produce. Since the function is a rational function, we can use the concept of horizontal asymptotes to determine the range. The horizontal asymptote of a rational function is the value that the function approaches as x approaches infinity. In this case, the horizontal asymptote is $y=2$. Therefore, the range of function B is all real numbers except $y=2$.

Analyzing Function C: $f(x)=\frac{x-3}{x+2}$

Finally, let's analyze function C: $f(x)=\frac{x-3}{x+2}$. To determine the domain of this function, we need to find the values of x that make the denominator equal to zero. In this case, the denominator is $x+2$, which equals zero when $x=-2$. Therefore, the domain of function C is all real numbers except $x=-2$.

To determine the range of function C, we need to find the values of y that the function can produce. Since the function is a rational function, we can use the concept of horizontal asymptotes to determine the range. The horizontal asymptote of a rational function is the value that the function approaches as x approaches infinity. In this case, the horizontal asymptote is $y=-1$. Therefore, the range of function C is all real numbers except $y=2$.

Conclusion

In conclusion, we have analyzed three functions and determined their domains and ranges. Function A has a domain of $x \neq -3$ and a range of $y \neq 2$. Function B has a domain of $x \neq -3$ but a range of $y \neq 2$. Function C has a domain of $x \neq -2$ and a range of $y \neq 2$. Therefore, the correct answer is function A.

References

• [1] "Domain and Range of Functions." Khan Academy, Khan Academy, www.khanacademy.org/math/algebra/x2f6b7d7f7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d7d