Problem 9
Question
Express the function as a composition of two simpler functions. $$ y=\sqrt{x^{2}+1} $$
Step-by-Step Solution
Verified Answer
Question: Express the function $$y=\sqrt{x^2+1}$$ as a composition of two simpler functions.
Answer: The given function $$y=\sqrt{x^2+1}$$ can be expressed as a composition of two simpler functions $$f(x)=x^2+1$$ and $$g(x)=\sqrt{x}$$, such that $$y=g(f(x))=\sqrt{x^2+1}$$.
1Step 1: Identify the two simpler functions
The given function is $$y=\sqrt{x^{2}+1}$$. In order to express this as a composition of two simpler functions, we can consider the following two functions:$$
f(x)=x^2+1 \\
g(x)=\sqrt{x}
$$
Notice how these two functions are simpler than the given function and it appears that their composition might result in the original function. Let's proceed to confirm this.
2Step 2: Find the composition of the two functions
Now we will find the composition of these two functions. We are aiming to find the composition $$g(f(x))$$ such that it results in the original function $$y=\sqrt{x^{2}+1}$$. Let's find out if these two functions can form such a composition:
$$
g(f(x)) = g(x^2+1)
$$
Since the function $$g(x)=\sqrt{x}$$, we just need to substitute $$x^2+1$$ into $$g(x)$$ to obtain the composition:
$$
g(f(x)) = \sqrt{x^2+1}
$$
3Step 3: Verify the composition
The composition of our two simpler functions, $$g(f(x))$$, is indeed$$
g(f(x))=\sqrt{x^{2}+1}
$$
This is the same as the original function $$y=\sqrt{x^{2}+1}$$, which confirms that the composition of the simpler functions, $$f(x)$$ and $$g(x)$$, results in the given function $$y$$.
Key Concepts
Square Root FunctionQuadratic FunctionMathematical Notation
Square Root Function
The square root function is often denoted as \(g(x) = \sqrt{x}\). It is the inverse of squaring a number. The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\). For example, the square root of 9 is 3, because \(3 \times 3 = 9\). The domain of the square root function is restricted to non-negative numbers because the square root of a negative number is not defined within the set of real numbers. The range of this function is also non-negative, meaning it only outputs non-negative numbers.
- Special Property: The function always returns the non-negative root, known as the principal square root.
- Graph Characteristics: The graph of \(g(x) = \sqrt{x}\) is a gentle curve rising from the origin, moving rightward along positive \(x\)-values.
Quadratic Function
A quadratic function is expressed in the standard form \(f(x) = ax^2 + bx + c\). In this exercise, we considered the simpler quadratic function \(f(x) = x^2+1\). The general form uses a squared term as its highest degree, making it a second-degree polynomial function.
- Nature of Quadratics: These functions often produce parabolic graphs. The direction of the parabola (opening upwards or downwards) depends on the sign of the coefficient \(a\). For instance, if \(a > 0\), the parabola opens upwards.
- Vertex: The point where the parabola changes direction is its vertex, and it can either be a maximum or a minimum point.
Mathematical Notation
Mathematical notation is a set of symbols and conventions used to communicate mathematical ideas clearly and succinctly. In the exercise, we use various notations:
- Function Notation: Functions are written as \(f(x)\), which represents the output of the function \(f\) for an input \(x\).
- Composition of Functions: Notated as \(g(f(x))\), this represents the application of function \(g\) to the result of function \(f\) applied to \(x\).
- Square Root: Denoted by \(\sqrt{x}\), indicating the principal square root or the non-negative root.
Other exercises in this chapter
Problem 9
In Exercises \(9-12,\) show that composing the functions in either order gets us back to where we started. $$ y=7 x-5 \text { and } x=\frac{y+5}{7} $$
View solution Problem 9
Find (a) The domain. (b) The range. $$ y=\sqrt{x+1} $$
View solution Problem 10
Show that composing the functions in either order gets us back to where we started. $$ y=8 x^{3} \text { and } x=\sqrt[3]{\frac{y}{8}} $$
View solution Problem 10
Find (a) The domain. (b) The range. $$ y=\frac{1}{x-2} $$
View solution