Problem 29 - Precalculus

Question

Use \(f(x)=-2 x, g(x)=\sqrt{x}\) and \(h(x)=|x|\) to find and simplify expressions for the following functions and state the domain of each using interval notation. $$(f \circ h \circ g)(x)$$

Step-by-Step Solution

Answer

The expression is \(-2 \sqrt{x}\) with domain \([0, \infty)\).

Step 1: Break Down the Composition

The function asked is \((f \circ h \circ g)(x) = f(h(g(x)))\). This means we first apply \(g\) to \(x\), then apply \(h\) to the result, and finally apply \(f\).

Step 2: Apply Function g(x)

The function \(g(x) = \sqrt{x}\). Thus, applying \(g\) to \(x\) gives \(g(x) = \sqrt{x}\). The domain of \(g(x)\) is \([0, \infty)\) because \(x\) must be non-negative for the square root to be defined.

Step 3: Apply Function h(x)

Now, apply \(h\) to \(g(x)\): \(h(g(x)) = h(\sqrt{x}) = |\sqrt{x}| = \sqrt{x}\), since the square root of a non-negative number is also non-negative, the absolute value doesn't change it. The domain remains \([0, \infty)\).

Step 4: Apply Function f(x)

Finally, apply \(f\) to \(h(g(x))\): \(f(h(g(x))) = f(\sqrt{x}) = -2\sqrt{x}\). The expression becomes \(-(2 \sqrt{x})\) with the domain of \([0, \infty)\) from previous steps.

Step 5: Write the Final Expression with Domain

The simplified expression for \((f \circ h \circ g)(x)\) is \( -2 \sqrt{x} \). The domain of this composed function is \([0, \infty)\).

Key Concepts

Domain of a function

The domain of a function refers to all the possible input values (usually represented by "x") that a function can handle without resulting in any undefined or problematic situation. To find the domain, we must consider the characteristics and limitations of each function involved, particularly if they involve operations like division by zero or taking square roots of negative numbers. In the case of the composition \((f \circ h \circ g)(x)\), the functions involved are

• \(g(x) = \sqrt{x}\),
• \(h(x) = |x|\), and
• \(f(x) = -2x\).

The domain of \(g(x)\) is \([0, \infty)\) because square roots require non-negative numbers. This restriction carries over through the composition, because each subsequent function relies on the output of the previous one. Hence, the domain of the entire composed function \((f \circ h \circ g)(x)\) is also \([0, \infty)\). This ensures that all operations within the composition work smoothly without undefined behaviors.

Square root function

A square root function like \(g(x) = \sqrt{x}\) operates under the principle that it can only take inputs that are non-negative. This means that the function only considers inputs \(x\geq 0\). It transforms the input value into the number which, when squared, gives back the original number \(x\). For example,

• \(\sqrt{9} = 3\), because \(3^2 = 9\).

This property of taking square roots restricts the domain to \([0, \infty)\) because attempting to find the square root of a negative number would result in complex numbers, which are not handled in the standard real number set.
In the function composition \((f \circ h \circ g)(x)\), the presence of \(g(x) = \sqrt{x}\) essentially sets the boundary for all following operations. It prevents negative inputs, ensuring that all subsequent manipulations by \(h\) and \(f\) remain valid and simple.

Absolute value function

The absolute value function, denoted as \(h(x) = |x|\), is an operation that transforms any real number into its non-negative counterpart. The function is straightforward:

• If \(x\) is positive or zero, \(|x| = x\).
• If \(x\) is negative, \(|x| = -x\).

This means that the absolute value function does not truly change non-negative inputs. It "checks" if the input is negative and, if so, flips the sign.
In the composition \((f \circ h \circ g)(x)\), we observe \(h(g(x)) = |\sqrt{x}| = \sqrt{x}\). Since \(\sqrt{x}\) is always non-negative for \(x \geq 0\), the absolute value has no effect on \(\sqrt{x}\). Thus, while it provides flexibility for negative inputs (by ensuring they become non-negative in general cases), it remains neutral when applied to the output of a square root, maintaining the function's domain as \([0, \infty)\).