Problem 34 - Intermediate Algebra

Question

Determine the indicated functional values. (Objective 2 ) If \(f(x)=\sqrt{x+6}\) and \(g(x)=3 x-1\), find \((f \circ g)(-2)\) and \((g \circ f)(-2)\).

Step-by-Step Solution

Answer

\((f \circ g)(-2)\) is undefined; \((g \circ f)(-2) = 5.\)

Step 1: Understand the Composition of Functions

The notation \((f \circ g)(x)\) represents the composition of two functions, which means we apply function \(g\) to \(x\), then apply function \(f\) to the result of \(g(x)\). Similarly, \((g \circ f)(x)\) means first apply \(f\) to \(x\), then \(g\) to the result of \(f(x)\).

Step 2: Calculate \(g(-2)\)

Substitute \(-2\) into the function \(g(x)\).\[g(-2) = 3(-2) - 1 = -6 - 1 = -7.\]

Step 3: Calculate \(f(g(-2))\) or \(f(-7)\)

Now, using the result from Step 2, substitute \(-7\) into the function \(f(x)\).\[f(-7) = \sqrt{-7 + 6} = \sqrt{-1}.\]Since the square root of a negative number is not a real number, \(f(-7)\) is undefined in the set of real numbers.

Step 4: Calculate \(f(-2)\)

Substitute \(-2\) into the function \(f(x)\).\[f(-2) = \sqrt{-2 + 6} = \sqrt{4} = 2.\]

Step 5: Calculate \(g(f(-2))\) or \(g(2)\)

Using the result from Step 4, substitute \(2\) into the function \(g(x)\).\[g(2) = 3 \times 2 - 1 = 6 - 1 = 5.\]

Step 6: Conclusion

From the steps above, we determined that \((f \circ g)(-2)\) is undefined, and \((g \circ f)(-2) = 5.\)

Key Concepts

Functional Values

When dealing with functional values, we focus on what happens when a particular input value is placed into a function. It is important to understand that each function has specific rules that dictate how outputs are calculated from inputs. For example, in our exercise, we have two functions, \( f(x) = \sqrt{x+6} \) and \( g(x) = 3x - 1 \). Here, the functional value depends heavily on the input provided and the nature of the function.

• The function \( f(x) = \sqrt{x+6} \) transforms input \( x \) by first adding 6, then taking the square root of the result.
• The function \( g(x) = 3x - 1 \) transforms input \( x \) by multiplying it by 3, and subtracting 1 from the outcome.

When solving for functional values, careful consideration is needed to abide by the operations indicated by the function expressions.

Real Numbers

Real numbers encompass all the numbers that can be found on the number line. This includes both positive and negative numbers, fractions, and most importantly, numbers like \( \pi \) and \( e \), which cannot be expressed as fractions. In the realm of functions, especially those involving radicals and other algebraic concepts, understanding real numbers is crucial.

• A number is classified as a real number if it exists on the continuous number line, meaning it can be rational or irrational.
• This exercise assumes that all results should ideally be real numbers unless specified otherwise.

It's essential to know that operations like taking square roots require caution. For instance, square roots of negative numbers fall outside the set of real numbers under typical circumstances, leading to special cases like undefined results.

Undefined in Real Numbers

Sometimes, when evaluating functions, the result may not exist within the constraints of real numbers. Such cases are what we refer to as "undefined in real numbers." This scenario often occurs with operations like division by zero or taking the square root of a negative number.

• In our exercise, evaluating \( f(g(-2)) \) involves attempting to take a square root of \( -1 \), which is not defined among real numbers.
• Whenever a step produces an output that doesn’t fit into the system of real numbers, it’s marked as 'undefined in real numbers.'

Recognizing when a function or mathematical operation shifts out of the real number set is pivotal in solving math problems. This knowledge helps prevent errors and guides you in understanding the function's boundaries.