Problem 61 - Precalculus Essentials

Question

Find. a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=\sqrt{x}, g(x)=x-1$$

Step-by-Step Solution

Answer

a. \((f \circ g)(x)\) = \(\sqrt{x-1}\) b. \((g \circ f)(x)\) = \((\sqrt{x})-1\) c. \((f \circ g)(2)\) = 1 d. \((g \circ f)(2)\) = \((\sqrt{2})-1\)

Step 1 - Find (f ⚪ g)(x)

To find \((f \circ g)(x)\) substitute \(g(x)\) into function \(f(x)\). Here, \(g(x)=x-1\) and \(f(x)=\sqrt{x}\). So, \((f \circ g)(x)\) = \(f(g(x))\) = \(\sqrt{x-1}\)

Step 2 - Find (g ⚪ f)(x)

To find \((g \circ f)(x)\), substitute \(f(x)\) into function \(g(x)\). Here, \(f(x)=\sqrt{x}\) and \(g(x)=x-1\). So, \((g \circ f)(x)\) = \(g(f(x))\) = \((\sqrt{x})-1\)

Step 3 - Find (f ⚪ g)(2)

To find \((f \circ g)(2)\), substitute 2 into \(g(x)\), then take that result and substitute it into \(f(x)\). Here, \(g(2)=2-1=1\). Therefore, \((f \circ g)(2)\) = \(f(g(2))\) = \(f(1)\) = \(\sqrt{1}=1\)

Step 4 - Find (g ⚪ f)(2)

To find \((g \circ f)(2)\), substitute 2 into \(f(x)\), then take that result and substitute it into \(g(x)\). Here, \(f(2)=\sqrt{2}\). Therefore, \((g \circ f)(2)\) = \(g(f(2))\) = \(g(\sqrt{2})\) = \((\sqrt{2})-1\)

Key Concepts

Understanding the Square Root Function

The square root function is a fundamental concept in mathematics. It’s denoted by the symbol "\(\sqrt{x}\)" and represents a value that, when squared, gives the original number, \(x\).
This function is particularly important because it has unique properties:

• It is only defined for non-negative values of \(x\) in the real number system, meaning \(x\) must be zero or positive.
• Its graph is half of a parabola, opening upwards, starting at the origin \((0,0)\).
• The square root function increases monotonically, meaning it never decreases.

Using the square root function, you can solve for various values and manipulate expressions that are squared, tying in closely with quadratic equations and geometry. Throughout this composition, we are focusing on integrating this function within another, aiding in better analytical and problem-solving abilities.

Exploring Substitution in Functions

Substitution in functions is a crucial method for composing or combining different functions. In function composition, you essentially replace the entire variable \(x\) in one function with another function.
This operation is written as \((f \circ g)(x)\), which means that you first apply function \(g\) to \(x\), and then apply function \(f\) to the result from \(g(x)\).
Function composition involves step-by-step substitution:

• First, substitute \(x\) in \(g(x)\), thereby finding the value of \(g(x)\).
• Next, take the result from \(g(x)\) and substitute it into \(f(x)\).
• This yields the composite function \(f(g(x))\).

Such substitution is handy in various mathematical and real-world applications when dealing with sequences of operations or complex formulas that need breakdown. This approach simplifies the way functions influence each other through their transformations.

Mastering Evaluation of Functions

Evaluation of functions is applying function rules to specific inputs to find their output. This is a key aspect of function analysis. To evaluate a function, you substitute a given value for the variable, and then perform the operations defined by the function.
Let's break this down further:

• Substitute the given input into the function equation.
• Follow through with any arithmetic operations involved.
• The result is the output or "value of the function" at the given input.

For example, evaluating \((f \circ g)(2)\) involves substituting 2 into \(g\), resulting in 1, and then finding \(f(1)\), which is \(\sqrt{1}\) equalling 1. Similarly, for \((g \circ f)(2)\), substitute 2 into \(f\) to get \(\sqrt{2}\), and then evaluate \(g(\sqrt{2}) = \sqrt{2} -1\). Evaluation ensures accuracy and enables us to predict values efficiently within a problem context. It’s an essential tool for verifying results and interpreting function behaviors.