Domain Of F(x) = Sqrt(5x-5)+1: Which Inequality?

When we're working with functions, one of the first things we often want to figure out is its domain. The domain is essentially the set of all possible input values (the 'x' values) for which the function is defined and produces a real number output. For the function $f(x)=\sqrt{5x-5}+1$, we need to be a bit careful because of that square root. You see, you can't take the square root of a negative number and get a real number result. This means the expression inside the square root, $5x-5$, must be greater than or equal to zero. This is the core principle we use to determine the domain of functions involving square roots. So, we set up an inequality: $5x-5 \geq 0$. This inequality will guide us to find all the valid 'x' values that can be plugged into our function without causing any mathematical problems. Let's dive deeper into why this inequality is the key and how to solve it to fully understand the domain of $f(x)$.

Understanding the Constraint: Why $5x-5 \geq 0$ is Crucial

The domain of a function represents all the possible 'x' values that make sense for that function. For $f(x)=\sqrt{5x-5}+1$, the most critical part is the square root. In the realm of real numbers, the square root operation is only defined for non-negative numbers (zero or positive numbers). If we try to input an 'x' value that makes the expression inside the square root negative, we'll end up with an imaginary number, and typically, when we talk about the domain of a function in introductory calculus or algebra, we're concerned with real-valued outputs. Therefore, the expression under the radical sign, which is $5x-5$, must satisfy the condition of being greater than or equal to zero. This is not just an arbitrary rule; it's a fundamental property of square roots in the real number system. Any 'x' that violates this condition will lead to an undefined result for $f(x)$ in the real number system. So, to find the domain, our primary focus must be on solving the inequality $5x-5 \geq 0$. This inequality acts as a gatekeeper, allowing only those 'x' values that result in a non-negative radicand.

Analyzing the Given Options

Let's look at the options provided to find the inequality used to determine the domain of $f(x)=\sqrt{5x-5}+1$:

A. $5x-4 \geq 0$: This inequality seems to be a slight misinterpretation. While it involves 'x', it doesn't directly address the expression inside the square root. The '+1' outside the square root doesn't affect the condition under the square root. So, this is likely incorrect.

B. $\sqrt{5x-5}+1 \geq 0$: This inequality states that the entire function's output must be greater than or equal to zero. While it's true that the square root part itself will always be non-negative (so the whole function will likely be $\geq 1$), this inequality doesn't help us find the domain. The domain is about what 'x' values are allowed into the function, not about the range of the function's output. The constraint comes from inside the square root, not from the overall function's value.

C. $5x \geq 0$: This is a simplified version of the expression inside the square root, but it misses the '-5'. It's like saying the 'x' part alone needs to be non-negative, ignoring the constant term that shifts the required value of 'x'. This is also incorrect.

D. $5x-5 \geq 0$: This inequality directly targets the expression inside the square root. As we've discussed, for the square root to yield a real number, its radicand (the expression inside) must be non-negative. This inequality perfectly captures that requirement. Therefore, this is the correct inequality to use for finding the domain of $f(x)=\sqrt{5x-5}+1$.

Solving the Inequality to Find the Domain

Now that we've identified $5x-5 \geq 0$ as the correct inequality for determining the domain, let's solve it. Our goal is to isolate 'x' to understand the range of values it can take.

1. Add 5 to both sides: To get the 'x' term by itself, we add 5 to both sides of the inequality: $5x - 5 + 5 \geq 0 + 5$ $5x \geq 5$

2. Divide by 5: Now, to find the value of 'x', we divide both sides by 5: $\frac{5x}{5} \geq \frac{5}{5}$ $x \geq 1$

So, the inequality $5x-5 \geq 0$ simplifies to $x \geq 1$. This means that the function $f(x)=\sqrt{5x-5}+1$ is defined for all real numbers 'x' that are greater than or equal to 1. Any 'x' value less than 1 would make the expression inside the square root negative, leading to an undefined real number result.

Visualizing the Domain

We can visualize this domain on a number line. We start at the number 1 and shade everything to the right, indicating all numbers greater than or equal to 1. We use a closed circle at 1 to show that 1 itself is included in the domain. In interval notation, this domain is written as $[1, \infty)$. This interval represents all real numbers from 1 up to positive infinity, inclusive of 1.

Why Other Inequalities Don't Work

Let's reiterate why the other options are incorrect by considering what happens if we use them:

• If we used $5x-4 \geq 0$: Solving this would give $5x \geq 4$, so $x \geq \frac{4}{5}$. If we input $x = \frac{4}{5}$ into our original function, we get $f(\frac{4}{5}) = \sqrt{5(\frac{4}{5})-5}+1 = \sqrt{4-5}+1 = \sqrt{-1}+1$. This results in an imaginary number, so this inequality does not correctly define the domain.

• If we used $\sqrt{5x-5}+1 \geq 0$: As mentioned earlier, this concerns the output of the function. The square root term $\sqrt{5x-5}$ is always non-negative by definition (for real numbers). Adding 1 to it means the entire function's output will always be 1 or greater. So, the inequality $\sqrt{5x-5}+1 \geq 0$ is always true for any 'x' in the domain of the square root. It doesn't impose any new restriction to find the domain itself.

• If we used $5x \geq 0$: Solving this gives $x \geq 0$. If we input $x=0$ into our function, we get $f(0) = \sqrt{5(0)-5}+1 = \sqrt{-5}+1$. Again, this results in an imaginary number. So, this inequality is too broad and includes values of 'x' that are not valid for the function.

Conclusion

For the function $f(x)=\sqrt{5x-5}+1$, the critical condition for its domain stems from the requirement that the expression under the square root must be non-negative. This leads directly to the inequality $5x-5 \geq 0$. By solving this inequality, we find that $x \geq 1$, which means the domain of the function is all real numbers greater than or equal to 1. The correct inequality used to find the domain is therefore D. $5x-5 \geq 0$. Understanding the properties of square roots is fundamental to correctly identifying and calculating the domain of such functions.

For further reading on functions and their domains, you can explore resources from Khan Academy.