Problem 55 - 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. $$(g \circ h)(-3)$$

Step-by-Step Solution

Answer

The composition function \((g \circ h)(-3)\) of two functions gives 3.

Step 1: Identify the function forms

The function forms are given as \(f(x)=x^{2}+x, g(x)=\sqrt{x},\) and \(h(x)=-3 x\). But in this particular exercise, only functions \(g\) and \(h\) are needed. So, we take \(g(x)=\sqrt{x}\) and \(h(x)=-3x\).

Step 2: Calculate the value of \(h(-3)\)

Substitute the value of \(x\) as \(-3\) in function \(h(x)\) \[h(-3) = -3*(-3)\] \[ h(-3) = 9 \]

Step 3: Compute the composition function \((g \circ h)(-3)\)

\((g \circ h)(-3)\) will be the function \(g\) evaluated at \(h(-3)\). Which would be \(g[h(-3)]\). We have already calculated that \(h(-3) = 9\). Substituting this in function \(g(x)\), we will get, \[g[h(-3)] = g(9) = \sqrt{9}\]

Step 4: Compute the value of \(g(9)\)

Now simply substitute \(9\) into function \(g\), that gives \[\sqrt{9} = 3\]

Key Concepts

Function Evaluation

Function evaluation is a process of determining the output of a function for a particular input. It's akin to following a recipe, where the input is your ingredient, and the function itself represents the instructions for what to do with that ingredient. This process is critical in understanding the nature of functions and how they behave.
When you are given a function, like in our example function h(x) = -3x, evaluating this function for a specific value involves substituting that value into the equation wherever the variable x appears. For instance, evaluating h(-3) means we replace every instance of x with -3, leading to the calculation h(-3) = -3 * (-3), which simplifies to h(-3) = 9.
Understanding function evaluation is a fundamental skill in algebra and calculus, as it lays the groundwork for more complex operations such as compositions, transformations, and even calculus-based concepts like limits and derivatives.

Square Roots

The square root function is an essential concept in mathematics, dealing with finding a number that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3, which is written as \(\sqrt{9}=3\), because \(3 \times 3 = 9\). In our exercise, function g(x) is described as g(x) = \sqrt{x}, meaning that for any input x, the output is the square root of x.
When we encounter square roots in function operations, we're often dealing with positive square roots, also known as the principal square root. It's important to remember that every positive number actually has two square roots: one positive and one negative. But in most cases, especially when we are dealing with function evaluations, we are interested in the principal, or non-negative, square root.

Function Operations

Function operations include addition, subtraction, multiplication, division, and composition of functions. Among these, function composition is a significant operation that involves applying one function to the results of another.
In the notation \((g \circ h)(x)\), the circle represents composition, meaning that function g is applied to the output of function h. In our exercise, we first evaluated h(-3) and then applied function g to the result, following the order inside-out, first h, then g. This 'inside-out' approach is crucial for correctly performing function compositions.
It's like a nesting doll scenario: the output of the inner doll (function h) becomes the input for the outer doll (function g). Understanding this order is vital as reversing the composition order can lead to different results, and it’s key to solving many algebraic problems involving functions.