Problem 54 - Brief Applied Calculus

Question

For each pair of functions \(f(x)\) and \(g(x)\), find a. \(f(g(x))\) b. \(g(f(x))\) and c. \(f(f(x))\) $$ f(x)=x^{2}+1 ; g(x)=x-\sqrt{x} $$

Step-by-Step Solution

Answer

\(f(g(x)) = x^2 - 2x\sqrt{x} + x + 1\), \(g(f(x)) = x^2 + 1 - \sqrt{x^2 + 1}\), \(f(f(x)) = x^4 + 2x^2 + 2\).

Step 1: Understand the Functions

We have two functions: \(f(x) = x^2 + 1\) and \(g(x) = x - \sqrt{x}\). We need to compute three compositions: \(f(g(x))\), \(g(f(x))\), and \(f(f(x))\).

Step 1: Calculate \(f(g(x))\)

We want to find \(f(g(x))\). Substitute \(g(x) = x - \sqrt{x}\) into the function \(f(x)\). So, \(f(g(x)) = f(x - \sqrt{x})\).\\[f(x - \sqrt{x}) = (x - \sqrt{x})^2 + 1 \] \\[= (x^2 - 2x\sqrt{x} + x) + 1 \] \\[= x^2 - 2x\sqrt{x} + x + 1 \] \So, \(f(g(x)) = x^2 - 2x\sqrt{x} + x + 1\).

Step 2: Calculate \(g(f(x))\)

We need to find \(g(f(x))\). Substitute \(f(x) = x^2 + 1\) into \(g(x)\). So, \(g(f(x)) = g(x^2 + 1)\).\\[g(x^2 + 1) = (x^2 + 1) - \sqrt{x^2 + 1} \] \Hence, \(g(f(x)) = x^2 + 1 - \sqrt{x^2 + 1}\).

Step 3: Calculate \(f(f(x))\)

Now, we find \(f(f(x))\) by substituting \(f(x) = x^2 + 1\) into itself.\\[f(f(x)) = f(x^2 + 1) = (x^2 + 1)^2 + 1 \] \\[= (x^4 + 2x^2 + 1) + 1 \] \\[= x^4 + 2x^2 + 2\] \Thus, \(f(f(x)) = x^4 + 2x^2 + 2\).

Key Concepts

Nested Functions

When working with nested functions, you're dealing with the idea of placing one function inside another. It's like putting a series of Russian dolls one inside the other. For example, if we have two functions, let's say \( f(x) = x^2 + 1 \) and \( g(x) = x - \sqrt{x} \), we can create new functions by composition, such as \( f(g(x)) \), \( g(f(x)) \), and \( f(f(x)) \). Each of these compositions involves taking the output of one function and using it as the input for another function.

Here's how it works:

• For \( f(g(x)) \), you first compute \( g(x) \) and then substitute that entire expression wherever you see \( x \) in \( f(x) \).
• For \( g(f(x)) \), you first evaluate \( f(x) \) and then input that into \( g(x) \), replacing \( x \) with \( f(x) \).
• \( f(f(x)) \) requires you to use \( f(x) \) as the argument for \( f(x) \) again, effectively applying the function twice in succession.

This process of nesting helps in understanding how complex functions build on simpler ones.

Polynomial Functions

Polynomial functions are an essential part of algebra and calculus. They are formed from sums of terms, each consisting of a variable raised to a non-negative integer power and multiplied by a coefficient, like \( f(x) = x^2 + 1 \). Here, the function is quadratic, meaning its highest exponent is 2.

A polynomial function can be easily manipulated with operations such as addition, subtraction, and multiplication. These operations preserve the polynomial structure, meaning you'll end up with another polynomial. For instance:

• Adding \( f(x) + f(x) \) simply doubles the coefficients of \( f(x) \).
• Squaring \( f(x) \) to find \( f(f(x)) \) involves expanding \((x^2 + 1)^2\), resulting in a new polynomial with terms of higher degree.

Polynomials are smooth and continuous, making them a crucial tool for modeling real-world phenomena, like projectile motion or the shape of a roller coaster.

Radical Functions

Radical functions include roots, such as square roots and cubics. These functions involve terms with a radical, like \( g(x) = x - \sqrt{x} \). The presence of a radical term introduces new considerations during calculations.

Here are a few characteristics of radical functions:

• Radicals can limit the domain of a function, since you typically cannot take the square root of a negative number (in real numbers).
• Combining radical functions with other types, like polynomial functions in compositions, can lead to interesting results. For example, in \( g(f(x)) = x^2 + 1 - \sqrt{x^2 + 1} \), the polynomial term \( x^2 + 1 \) is still present but is modified by the radical term.
• Radicals often produce equations that are harder to solve analytically, hence requiring special methods or numerical approximations.

Understanding these properties helps in identifying when radicals may impact calculations, especially in function compositions.