Problem 74 - Precalculus Essentials

Question

Find; a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=x^{2}+1, g(x)=\sqrt{2-x}$$

Step-by-Step Solution

Answer

a. \((f \circ g)(x) = 3 - x\)\nb. The domain of \(f \circ g\) is \(x \leq 2\)

Step 1: Evaluate the composite function

For \(f \circ g\), evaluate function \(f\) at function \(g\). Thus, \(f(g(x)) = f(\sqrt{2-x}) = (\sqrt{2-x})^2 + 1 = 2 - x + 1 = 3 - x.\)

Step 2: Determine the domain of g and f(g)

The domain of the function \(g(x) = \sqrt{2-x}\) is the set of all \(x\) such that \(2-x \geq 0\). Solving this inequality, we get \(x \leq 2\). But since we are dealing with \(f \circ g\), we also need to consider the composite function. The domain of \(f(g(x)) = 3 - x\) is all real numbers. However, as \(g\) is part of \(f \circ g\), we must take into account its domain as well, hence the domain of \(f \circ g\) is \(x \leq 2\)

Key Concepts

Function Composition

Function composition is an operation that takes two functions and combines them to form a new function. It is denoted by \((f \circ g)(x)\), which means you apply the function \(g\) to \(x\) first, and then apply the function \(f\) to the result. Think of it as a process akin to a conveyer belt: the output from \(g(x)\) becomes the input for \( f(x)\). This method allows for chaining of functions and solving more complex mathematical problems.
For example, if \(f(x) = x^2 + 1\) and \(g(x) = \sqrt{2-x}\), the composition \((f \circ g)(x)\) is found by substituting \(g(x)\) into \(f\). So, \((f(g(x)) = f(\sqrt{2-x}))\), which turns into \((\sqrt{2-x})^2 + 1 = 2 - x + 1\), thus simplifying to \(3 - x\).
Composing functions is a powerful tool used in calculus and higher-level mathematics, allowing for the manipulation and transformation of equations across various problem contexts.

Domain of a Function

The domain of a function refers to all the possible input values (\(x\)-values) that a function can accept. It's essentially the list of numbers you can plug into the function without causing any issues like division by zero or taking square roots of negative numbers, which are not defined in the set of real numbers.
In the case of the composite function \(f \circ g\) from our example, we start by determining the domain of \(g(x) = \sqrt{2-x}\). The square root function requires non-negative values under the root, so we set the expression \(2-x \geq 0\), leading to \(x \leq 2\). This inequality tells us that \(x\) can be any real number up to and including 2.
Since the function \(f(x) = x^2 + 1\) is a polynomial, its domain is all real numbers. However, in \(f \circ g\), because the output of \(g\) is the input to \(f\), the domain must also respect the limits imposed by \(g\). Therefore, the domain of \(f \circ g\) remains \(x \leq 2\). This intersection of domains ensures the composite function is well-defined over all allowed \(x\)-values.

Square Root Function

The square root function, written as \(g(x) = \sqrt{x}\), extracts the square root of its input. It's one of the elementary functions and appears often in mathematics, especially in contexts requiring balance or symmetry, such as geometrical computations and natural growth models.
When dealing with square roots, it’s important to remember:

• Square roots are only defined for non-negative inputs in the realm of real numbers.
• The expression under the square root, known as the radicand, must be \(\geq 0\) for the square root to be defined.

In our scenario, \(g(x) = \sqrt{2-x}\) demands that the radicand \(2-x\) remain non-negative. This requirement restricts \(x\) to \(x \leq 2\), ensuring that you do not try to take the square root of a negative number, which would be undefined.
To maximize comprehension and problem-solving efficiency, always check the conditions imposed by square roots and adjust the possible values (domain) of your functions accordingly.