Domain Of (f O G)(x) With F(x)=2x And G(x)=sqrt(x)

Let's dive into the fascinating world of function composition and explore the domain of $(f \circ g)(x)$ when $f(x) = 2x$ and $g(x) = \sqrt{x}$. Understanding the domain of composite functions is a fundamental concept in mathematics, crucial for analyzing function behavior and solving various mathematical problems. The domain of a function refers to the set of all possible input values (x-values) for which the function is defined. When we compose two functions, say $f$ and $g$, to create a new function $(f \circ g)(x)$, which is equivalent to $f(g(x))$, we need to consider the restrictions imposed by both the inner function $g(x)$ and the outer function $f(x)$. This means that an input value $x$ must first be valid for $g(x)$, and then the output of $g(x)$ must be valid as an input for $f(x)$. In our specific case, $f(x) = 2x$ is a linear function, and its domain is all real numbers. However, $g(x) = \sqrt{x}$ is a square root function, and its domain is restricted to non-negative real numbers because the square root of a negative number is not a real number. Therefore, for $(f \circ g)(x) = f(g(x))$ to be defined, $x$ must be in the domain of $g$, and $g(x)$ must be in the domain of $f$. Since the domain of $g(x) = \sqrt{x}$ is $x \geq 0$, and the domain of $f(x) = 2x$ is all real numbers, we only need to ensure that $x$ is a valid input for $g(x)$. This leads us to conclude that the domain of $(f \circ g)(x)$ is the set of all $x$ such that $x \geq 0$. This aligns with option A, which states $x \geq 0$. Let's explore this further by first evaluating the composite function itself. $(f \circ g)(x) = f(g(x)) = f(\sqrt{x}) = 2\sqrt{x}$. Now, when we look at the resulting function $2\sqrt{x}$, it is evident that the expression under the square root, $x$, must be non-negative for the function to yield a real number output. Thus, the domain of $(f \circ g)(x)$ is indeed $x \geq 0$. This careful consideration of the inner function's domain is paramount when determining the domain of a composite function.

Understanding Function Composition

Function composition, denoted as $(f \circ g)(x)$, is a mathematical operation where the output of one function becomes the input of another. It's like a relay race for numbers! We start with an input $x$, pass it to the inner function $g$, get an output $g(x)$, and then use that output as the input for the outer function $f$, resulting in $f(g(x))$. The notation $(f \circ g)(x)$ is a concise way to represent this process. To determine the domain of such a composite function, we must adhere to two critical conditions. First, the input value $x$ must be within the domain of the inner function, $g(x)$. If $x$ is not a valid input for $g$, then $g(x)$ is undefined, and consequently, $f(g(x))$ cannot be evaluated. Second, the output of the inner function, $g(x)$, must be within the domain of the outer function, $f(x)$. Even if $x$ is a valid input for $g$, if the resulting $g(x)$ is not a valid input for $f$, then $f(g(x))$ remains undefined. In our specific problem, we have $f(x) = 2x$ and $g(x) = \sqrt{x}$. Let's analyze the domains of these individual functions. The function $f(x) = 2x$ is a simple linear function. Its graph is a straight line, and it accepts any real number as an input. Therefore, the domain of $f(x)$ is all real numbers, often represented as $(-\infty, \infty)$. On the other hand, $g(x) = \sqrt{x}$ is a square root function. The square root operation is only defined for non-negative numbers in the realm of real numbers. Taking the square root of a negative number results in an imaginary number, and typically, when discussing domains in introductory calculus and algebra, we are concerned with real-valued functions. Thus, the domain of $g(x)$ is all non-negative real numbers, which can be written as $[0, \infty)$ or $x \geq 0$. Now, let's apply the two conditions for the domain of $(f \circ g)(x) = f(g(x))$.

Condition 1: $x$ must be in the domain of $g(x)$. As we established, the domain of $g(x) = \sqrt{x}$ is $x \geq 0$. This means that any input $x$ we consider for $(f \circ g)(x)$ must be greater than or equal to zero.

Condition 2: $g(x)$ must be in the domain of $f(x)$. The output of $g(x)$ is $\sqrt{x}$. The domain of $f(x) = 2x$ is all real numbers. This means that any real number, including the outputs of $g(x)$ (which are always non-negative real numbers), can be plugged into $f(x)$. So, this second condition does not impose any additional restrictions beyond what's already required by the domain of $g(x)$.

Combining these conditions, the only restriction comes from the domain of the inner function $g(x)$. Therefore, the domain of $(f \circ g)(x)$ is $x \geq 0$. This corresponds to option A.

Step-by-Step Calculation of the Composite Function

To further solidify our understanding, let's explicitly calculate the composite function $(f \circ g)(x)$. The definition of function composition states that $(f \circ g)(x) = f(g(x))$. This means we take the entire expression for $g(x)$ and substitute it into the variable $x$ in the function $f(x)$.

We are given: $f(x) = 2x$ $g(x) = \sqrt{x}$

To find $f(g(x))$, we replace every instance of $x$ in $f(x)$ with the expression for $g(x)$: $f(g(x)) = 2(g(x))$ Now, substitute $g(x) = \sqrt{x}$: $f(g(x)) = 2(\sqrt{x})$ So, the composite function is $(f \circ g)(x) = 2\sqrt{x}$.

Now that we have the explicit form of the composite function, we can determine its domain by examining the expression $2\sqrt{x}$. For this expression to be defined in the set of real numbers, the term under the square root symbol must be non-negative. That is, $x$ must be greater than or equal to 0.

$x \geq 0$

This confirms our previous reasoning based on the domains of the individual functions. The domain of $(f \circ g)(x)$ is the set of all real numbers $x$ such that $x \geq 0$. This is represented by the interval $[0, \infty)$.

Let's consider why the other options are incorrect:

• B. $x \neq 0$: This option would imply that $x=0$ is excluded. However, $g(0) = \sqrt{0} = 0$, and $f(0) = 2(0) = 0$. So, $(f \circ g)(0) = 0$, which is perfectly valid. Therefore, $x=0$ must be included in the domain.

• C. $x \leq 0$: This option would mean that only non-positive numbers are allowed. If we tried to input a negative number, say $x = -4$, into $g(x) = \sqrt{x}$, we would get $\sqrt{-4}$, which is not a real number. Therefore, negative numbers are not in the domain of $g(x)$, and consequently, not in the domain of $(f \circ g)(x)$.

• D. all real numbers: This would mean that any real number can be inputted into $(f \circ g)(x)$. As we've seen, negative real numbers are not allowed because they are not in the domain of $g(x) = \sqrt{x}$.

Therefore, the only correct domain for $(f \circ g)(x)$ is $x \geq 0$.

Visualizing the Domains

To further grasp the concept, let's visualize the domains on a number line. The domain of $g(x) = \sqrt{x}$ starts at 0 and extends infinitely to the right. It includes 0.

Number line for $g(x)$ domain: [-----0-----1-----2-----3-----

Now, consider the domain of $f(x) = 2x$. It covers the entire number line:

Number line for $f(x)$ domain: -----(-inf)------------------(inf)-----

When we compose $(f \circ g)(x) = f(g(x))$, the input $x$ must first be valid for $g$. This means $x$ must be in the set $[0, \infty)$. Then, the output $g(x)$ (which is $\sqrt{x}$) must be valid for $f$. Since the domain of $f$ is all real numbers, any output from $g(x)$ is acceptable for $f$. Therefore, the restrictions on the domain of $(f \circ g)(x)$ are solely determined by the restrictions on the domain of $g(x)$. This is a key principle: the domain of a composite function $(f \circ g)(x)$ is the set of all $x$ in the domain of $g$ such that $g(x)$ is in the domain of $f$.

In our case: Domain of $g$: $x \geq 0$ Domain of $f$: All real numbers.

We need $x$ such that $x \geq 0$ AND $g(x)$ is a real number. Since $g(x) = \sqrt{x}$, $g(x)$ is a real number if and only if $x \geq 0$. Therefore, the combined condition is simply $x \geq 0$.

Conclusion: The Domain of $(f \circ g)(x)$

In conclusion, when $f(x) = 2x$ and $g(x) = \sqrt{x}$, the domain of the composite function $(f \circ g)(x)$ is determined by the restrictions imposed by the inner function, $g(x)$. Since $g(x) = \sqrt{x}$ requires its input $x$ to be non-negative ($x \geq 0$) for its output to be a real number, and the outer function $f(x) = 2x$ can accept any real number as input, the domain of $(f \circ g)(x)$ is limited by the domain of $g(x)$. Thus, the domain of $(f \circ g)(x)$ is all real numbers $x$ such that $x \geq 0$. This corresponds to Option A. It's a valuable lesson in understanding how the constraints of individual functions propagate through composition.

For further exploration into function composition and domains, you can visit Khan Academy's section on composite functions.