Problem 59 - Precalculus : Building Concepts and Connections

Question

In Exercises \(49-66,\) let \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\) Evaluate each of the following. $$(g \circ f)(2)$$

Step-by-Step Solution

Answer

\((g \circ f)(2) = \sqrt{6}\

Step 1: Evaluate Function f at x=2

First, evaluate the function \(f(x) = x^2 +x\) at the value x=2. Substitute x=2 into the function to get the value: \(f(2) = 2^2 +2 = 6\).

Step 2: Evaluate Function g at f(2)

Now substitute \(f(2)\) into the function \(g(x)\). Specifically, evaluate: \(g(f(2)) = g(6) = \sqrt{6}\).

Step 3: Simplify the Result

The function \(g(f(2))\) simplifies to \(\sqrt{6}\), which cannot be further simplified. Therefore, \((g \circ f)(2) = \sqrt{6}\).

Key Concepts

Evaluating Functions

Evaluating functions is a fundamental concept in algebra and calculus. It involves calculating the output of a function for a given input value. A function, like a machine, receives an input, processes it according to its rule, and provides an output.

• To evaluate a function, substitute the input value into the function's formula.
• Perform the arithmetic operations as defined by the function.

In the original exercise, we were tasked with evaluating a composite function involving two functions: \(f(x) = x^2 + x\) and \(g(x) = \sqrt{x}\). This required us to calculate \(f(2)\) first, which involved substituting \(2\) into \(f(x)\) to get the result of \(6\). Once we had \(f(2)\), we then substituted it into \(g(x)\), giving us \(g(6) = \sqrt{6}\). This final value, \(\sqrt{6}\), is the output when the composite function is evaluated at \(x = 2\).

Square Root Function

The square root function, represented as \(g(x) = \sqrt{x}\), is a mathematical function that pairs each non-negative number \(x\) with a non-negative number whose square is \(x\). It's essential to understand that:

• Square roots are defined only for non-negative inputs in the real number system.
• The output of \(\sqrt{x}\) can be seen as "the number that, when squared, returns the original \(x\)."

For instance, in the problem given, \(g(f(2))\) leads us to calculate \(g(6) = \sqrt{6}\). Even though \(\sqrt{6}\) is not a perfect square, it is still a legitimate number which can often appear in the result of evaluations of mathematical functions. Simplifying square roots further requires factoring, but \(6\) offers no simplification other than itself being a product of primes.

Quadratic Function

Quadratic functions are a type of polynomial function where the variable is squared, and it takes the form \(f(x) = ax^2 + bx + c\). They are best known for their characteristic U-shaped curves when graphed, called parabolas.

• In our exercise, \(f(x) = x^2 + x\), a form with coefficients \(a=1\), \(b=1\), and \(c=0\).
• This example is a simple quadratic function without a constant term, making it easier to evaluate.

When evaluating \(f(x)\) at certain values, you substitute the values into the equation. For example, substituting \(x = 2\), we computed \(f(2) = 2^2 + 2 = 6\). This is the step before substituting into another function, \(g(x)\), indicating the interplay between different function types in function compositions.

Function Notation

Function notation is a standardized way to represent functions and their evaluations. Using symbols like \(f(x)\), \(g(x)\), or \(h(x)\), you can convey complex mathematical ideas in a concise format.

• It helps distinguish between different functions and their variables.
• Allows simplification of mathematical expressions and makes communication clearer.

Function notation clarifies what operations will be performed. For example, in the exercise, the notation \((g \circ f)(x)\) indicates function composition, showing that function \(f\) should be evaluated first and the result will be used as the input for function \(g\). By following function notation, you can achieve accurate results efficiently, whether it's evaluating simple or complex functions.