+1252 Find domain and range of \(g(x) = \frac{x^3 + 11x - 2}{|x - 3| + |x + 1|}.\)
+2446 Let's first think of the domain of this particular function.
The only portion of this function that can affect the state of the domain is the denominator, so let's investigate it.
If there was a value for x such that \(|x-3|+|x+1|=0\), then a restriction for the domain of this function would exist. Let's see if there are any values for x such that this is true:
| \(|x-3|+|x+1|=0\) | I think the best way to solve this is to think about it logically. You could compute the sign changes, but I think there is a better way.
The absolute value signs always output a nonnegative result. In this equation, we are adding two nonnegative outputs together. There is only one way that the sum of two nonnegative numbers can equal 0. That way is 0+0=0. |
| \(|x-3|=0\\ x-3=0\\ x=3\) | I have determined that the only x-value that makes the first addend, \(|x-3|\), equal to 0 is when x=3. |
| \(|3+1|\stackrel{?}=0\\ 4\stackrel{?}=0\\ \text{false}\) | x=3 is the only candidate that can make this equation equal to 0. The second addend, |x+1|, evaluates to 4 when x=3. However, we determined earlier that |x+1| has to evaluate to 0 in order for solutions to exist. Therefore, there are no solutions, for the real number set at least. |
| \(\text{Domain}: x \in {\rm I\!R}\) | The conclusion from this investigation is that there are no restrictions in the domain. |
The nature of this function is generally one of a cubic function because there are no vertical asymptotes. Dividing by the addition of a sum of nonnegatives numbers does not really affect the range.
\(\text{Range}: x \in {\rm I\!R}\)
+2446 Let's first think of the domain of this particular function.
The only portion of this function that can affect the state of the domain is the denominator, so let's investigate it.
If there was a value for x such that \(|x-3|+|x+1|=0\), then a restriction for the domain of this function would exist. Let's see if there are any values for x such that this is true:
| \(|x-3|+|x+1|=0\) | I think the best way to solve this is to think about it logically. You could compute the sign changes, but I think there is a better way.
The absolute value signs always output a nonnegative result. In this equation, we are adding two nonnegative outputs together. There is only one way that the sum of two nonnegative numbers can equal 0. That way is 0+0=0. |
| \(|x-3|=0\\ x-3=0\\ x=3\) | I have determined that the only x-value that makes the first addend, \(|x-3|\), equal to 0 is when x=3. |
| \(|3+1|\stackrel{?}=0\\ 4\stackrel{?}=0\\ \text{false}\) | x=3 is the only candidate that can make this equation equal to 0. The second addend, |x+1|, evaluates to 4 when x=3. However, we determined earlier that |x+1| has to evaluate to 0 in order for solutions to exist. Therefore, there are no solutions, for the real number set at least. |
| \(\text{Domain}: x \in {\rm I\!R}\) | The conclusion from this investigation is that there are no restrictions in the domain. |
The nature of this function is generally one of a cubic function because there are no vertical asymptotes. Dividing by the addition of a sum of nonnegatives numbers does not really affect the range.
\(\text{Range}: x \in {\rm I\!R}\)
