Problem 53 - College Algebra with Modeling and Visualization

Question

Exercises \(53-56:\) Use the given \(f(x)\) and \(g(x)\) to evaluate each expression. \(f(x)=\sqrt{x+5}, \quad g(x)=x^{2}\) (a) \((f \circ g)(2)\) (b) \((g \circ f)(-1)\)

Step-by-Step Solution

Answer

(f \circ g)(2) = 3 and (g \circ f)(-1) = 4

Step 1: Understand Composition of Functions

The composition of functions \( (f \circ g)(x) \) means applying \(g(x)\), then applying \(f(x)\) to the result. Conversely, \( (g \circ f)(x) \) means applying \(f(x)\), then applying \(g(x)\) to the result. These concepts will guide the solution for both parts of the problem.

Step 2: Evaluate Part (a) \\( (f \circ g)(2) \\\)

To evaluate \( (f \circ g)(2) \), first evaluate \( g(2) \). Using \( g(x) = x^2 \), we find \( g(2) = 2^2 = 4 \). Next, substitute \( g(2) = 4 \) into \( f(x) = \sqrt{x+5} \), giving us \( f(4) = \sqrt{4+5} = \sqrt{9} = 3 \). Thus, \( (f \circ g)(2) = 3 \).

Step 3: Evaluate Part (b) \\( (g \circ f)(-1) \\\)

To evaluate \( (g \circ f)(-1) \), start by calculating \( f(-1) \). Using \( f(x) = \sqrt{x+5} \), calculate \( f(-1) = \sqrt{-1+5} = \sqrt{4} = 2 \). Then, substitute \( f(-1) = 2 \) into \( g(x) = x^2 \), yielding \( g(2) = 2^2 = 4 \). Therefore, \( (g \circ f)(-1) = 4 \).

Key Concepts

Function Evaluation

Function evaluation is the process of finding the output of a function given a specific input. To evaluate a function, you substitute the input value into the function's equation and perform the necessary operations to find the result. Understanding how to evaluate functions is critical in solving more complex problems involving functions, such as function composition.

Consider the simple functions provided in the exercise:

• For \( f(x) = \sqrt{x+5} \), evaluating involves substituting a given value for \(x\) and then calculating the square root of \(x+5\).
• For \( g(x) = x^2 \), it involves substituting a value into \(x^2\) to square it and find the answer.

Function evaluation is the foundation of function composition. It guides us through each step, helping us to derive the desired outcome swiftly.

Square Root Function

A square root function is a type of function that includes a square root in its formula, and is written as \( f(x) = \sqrt{x} \). The square root function typically has a restricted domain, meaning it can only take non-negative inputs to result in real numbers.

In this exercise, our square root function is \( f(x) = \sqrt{x+5} \). Here, we're looking at the square root of \(x+5\). This addition shifts the basic square root function horizontally.

• The domain of \( f(x) = \sqrt{x+5} \) consists of all \(x \) values for which \(x+5 \geq 0\), or \(x \geq -5\).
• Evaluating \(f(x)\) for inputs within this domain involves simple substitution and calculation of the square root.

The square root function graphically represents a curve that starts at a specific point determined by the domain, and rises gradually as \(x\) increases.

Quadratic Function

Quadratic functions are second-degree polynomials of the form \( g(x) = ax^2 + bx + c \). They graph into a parabolic shape, either opening upwards or downwards, depending on the sign of \(a\). These functions are instrumental in various mathematical applications.

In this case, we have a simpler version: \( g(x) = x^2 \). This basic quadratic function is symmetric and opens upwards from the origin. Key characteristics of this function include:

• A vertex located at the origin (0,0).
• A minimum value at \(x = 0\) for \(g(x)\), making the vertex the lowest point on the graph.

When you evaluate \(g(x)\), you substitute your specific value of \(x\) and calculate \(x^2\). In our exercise, evaluating \(g(2)\) leads us smoothly through composition by squaring the input \(2\), resulting in \(4\). This process is key in understanding how quadratic functions operate within compositions.