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What is the domain of \( \left(\frac{f}{g}\right)(x) \) ? \[ f(x)=-4 x+2 \] \( g(x)=5 x+2 \) All real numbers All real numbers except \( -\frac{2}{5} \) All real numbers except \( \frac{2}{5} \) All real numbers except \( -\frac{5}{2} \)

Ask by Blake Ross. in the United States
Feb 03,2025

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Answer

The domain of \(\frac{f}{g}(x)\) is all real numbers except \( -\frac{2}{5} \).

Solution

To determine the domain of the function \(\left(\frac{f}{g}\right)(x)\), where: \[ f(x) = -4x + 2 \quad \text{and} \quad g(x) = 5x + 2 \] we need to identify all real numbers \(x\) for which the function is defined. The function \(\frac{f}{g}(x)\) is defined as long as the denominator \(g(x)\) is not equal to zero, because division by zero is undefined. 1. **Find where \(g(x) = 0\):** \[ 5x + 2 = 0 \implies x = -\frac{2}{5} \] 2. **Domain of \(\frac{f}{g}(x)\):** The function \(\frac{f}{g}(x)\) is defined for all real numbers except where \(g(x) = 0\). Therefore, the domain excludes \(x = -\frac{2}{5}\). **Conclusion:** The domain of \(\frac{f}{g}(x)\) is all real numbers except \( -\frac{2}{5} \). **Answer:** All real numbers except \( -\frac{2}{5} \)

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To find the domain of \( \left(\frac{f}{g}\right)(x) \), we must consider where the function \( g(x) \) is not equal to zero, since division by zero is undefined. Now, we can set up the equation \( g(x) = 5x + 2 \) and solve for \( x \) when it equals zero: \[ 5x + 2 = 0 \] \[ 5x = -2 \] \[ x = -\frac{2}{5} \] Thus, the domain of \( \left(\frac{f}{g}\right)(x) \) is all real numbers except \( -\frac{2}{5} \). The correct answer is "All real numbers except \( -\frac{2}{5} \)." --- Now, if you ever find yourself grappling with domains, remember that the key players are the denominators! Just like your favorite superhero can't save the day if their powers are negated, your functions can't operate when division leads to zero. In practical scenarios, think about how domains apply in fields such as engineering, where understanding restrictions on variables is critical for designing safe structures. If a calculation suddenly requires an undefined value, it’s like trying to fit a square peg into a round hole—definitely not going to work!

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