In machine learning, the loss function used to train a model often combines several components that measure different aspects of performance.

Suppose we define:

• $f(x)$: The prediction error
• $g(x)$: A regularization term that penalizes overly complex models

Note that $x$ may represent several model parameters, but the idea of combining functions, i.e., adding terms that capture different effects, follows the same principle as in the single-variable case.

The resulting loss function balances accuracy (how well predictions match the observed data) with simplicity (how small the model parameters are):

$$L(x)=f(x)+\lambda \cdot g(x),$$

where $\lambda >0$ controls how strongly the regularization term influences the model.

In many real-world models, new relationships are created by combining existing quantities using arithmetic operations.

Suppose we define:

• $f(x)$: the temperature (in °C)
• $g(x)$: the humidity (in %)

A new function $h(x)$ can be defined to estimate a heat index (a perceived temperature) as follows:

$$h(x)=f(x)+0.1\cdot g(x)$$

Here, $h(x)$ is obtained by adding a weighted contribution from humidity to the temperature. Such combinations describe how different quantities together determine a result. In this case, both temperature and humidity contribute to the perceived heat.

In this example, we explore how subtraction and division affect the relationship between two functions. For this purpose, let

$$f(x)=x-1,\text{ and }g(x)=x^2-1.$$

We will now find and simplify both $(g-f)(x)$ and $\left(\frac{g}{f}\right)(x)$ to see how these operations transform the expressions.

First, subtract $f(x)$ from $g(x)$:

$$\begin{array}{rl}(g-f)(x)& =g(x)-f(x)\\ & =(x^2-1)-(x-1)\\ & =x^2-x\\ & =x(x-1)\end{array}$$

Then, divide $g(x)$ by $f(x)$:

$$\begin{array}{rl}\left(\frac{g}{f}\right)(x)& =\frac{g(x)}{f(x)}\\ & =\frac{x^2-1}{x-1}\\ & =\frac{(x+1)(x-1)}{x-1}\\ & =x+1,x\ne 1\end{array}$$

We can see that subtraction and division lead to very different results, i.e., $(g-f)(x)$ is a quadratic expression, while $\frac{g}{f}$ agrees with the linear expression $x+1$ only on its domain $\mathbb{R}\setminus \{1\}$. It is not the same function as $x+1$ on all of $\mathbb{R}$ because its graph has a hole at $x=1$.

Even though both start from the same $f$ and $g$, the way we combine them changes the type of function we obtain.

In this example, we explore how multiplication and subtraction affect the relationship between two functions. For this purpose, let

$$f(x)=x-1,\text{ and }g(x)=x^2-1$$

We will now find and simplify both $(f\cdot g)(x)$ and $(f-g)(x)$ to see how these operations transform the expressions.

First, multiply $f(x)$ and $g(x)$:

$$\begin{array}{rl}(f\cdot g)(x)& =f(x)\cdot g(x)\\ & =(x-1)(x^2-1)\\ & =x^3-x^2-x+1\end{array}$$

Then, subtract $g(x)$ from $f(x)$:

$$\begin{array}{rl}(f-g)(x)& =f(x)-g(x)\\ & =(x-1)-(x^2-1)\\ & =x-x^2\end{array}$$

Again, the two resulting functions are very different, i.e., $(f\cdot g)(x)$ is cubic, while $(f-g)(x)$ is quadratic.

Suppose we want to calculate how much electricity is used to cool a house on a particular day of the year. The electricity usage depends on the average indoor-outdoor temperature difference, which in turn depends on the average daily temperature outside.

Thus, we have two relationships:

• $E(T)$: describes the electricity (in kWh) required to maintain a desired indoor temperature for a given outdoor temperature $T$ (°C)
• $T(d)$: describes the average outdoor temperature (°C) on day $d$ of the year

For any given day $d$, the electricity use depends on the temperature, which itself depends on the day. We can therefore evaluate $E$ at the temperature given by $T(d)$:

$$E(T(d)).$$

This expression represents the electricity used on day $d$. For example, to find the electricity usage on the 10th day of the year, we first compute $T(10)$ and then use that value as the input to $E$. Thus, $E(T(10))$ gives the electricity required to cool the house on the 10th day of the year.

A quick calculation shows why the order of composition matters.

Using the following functions, find both $f(g(x))$ and $g(f(x))$ to determine whether composition is commutative.

$$f(x)=2x+1,g(x)=3-x$$

First, substitute $g(x)$ into $f(x)$:

$$\begin{array}{rl}f(g(x))& =2(3-x)+1\\ & =6-2x+1\\ & =7-2x\end{array}$$

Next, substitute $f(x)$ into $g(x)$:

$$\begin{array}{rl}g(f(x))& =3-(2x+1)\\ & =3-2x-1\\ & =2-2x\end{array}$$

Because $f(g(x))\ne g(f(x))$, we see that function composition is not commutative.

When decomposing a function, the goal is to identify an inner expression and an outer operation.

Express $h(x)=\sqrt{5-x^2}$ as the composition of two simpler functions.

We are looking for functions $f$ and $g$ such that

$$h(x)=(f\circ g)(x)=f(g(x))$$

To identify these functions, notice that $5-x^2$ appears inside the square root. This suggests the inner function produces $5-x^2$, and the outer function takes the square root of its input. Thus, we can define

$$g(x)=5-x^2\text{ and }f(x)=\sqrt{x}$$

We can verify our decomposition by recomposing the functions:

$$(f\circ g)(x)=f(g(x))=f(5-x^2)=\sqrt{5-x^2}$$

Therefore, $h(x)=(f\circ g)(x)$ with

$$g(x)=5-x^2,\text{ and }f(x)=\sqrt{x}$$

The domain of $h$ is $[-\sqrt{5},\sqrt{5}]$. On this domain, $g(x)\ge 0$, so its outputs are valid inputs for the square-root function $f:[0,\infty )\to [0,\infty )$.

For a more nested expression, we use the same idea but may group a larger inner expression first.

Express $h(x)=\frac{4}{3-\sqrt{4+x^2}}$ as the composition of two simpler functions.

We are looking for functions $f$ and $g$ such that

$$h(x)=(f\circ g)(x)=f(g(x))$$

Here, the expression $\sqrt{4+x^2}$ appears inside the denominator. We can treat that as the output of the inner function $g$, and then let the outer function $f$ operate on that result. Thus, we can define

$$g(x)=\sqrt{4+x^2}\text{ and }f(x)=\frac{4}{3-x}$$

We can verify our decomposition by recomposing the functions:

$$(f\circ g)(x)=f(g(x))=f\left(\sqrt{4+x^2}\right)=\frac{4}{3-\sqrt{4+x^2}}$$

Therefore, $h(x)=(f\circ g)(x)$ with

$$g(x)=\sqrt{4+x^2},\text{ and }f(x)=\frac{4}{3-x}$$

The denominator is zero when $\sqrt{4+x^2}=3$, which occurs at $x=\pm \sqrt{5}$. Therefore, the domain of $h$ is $\mathbb{R}\setminus \{-\sqrt{5},\sqrt{5}\}$.