Continuity and Differentiability - Algebra of continuous functions

Prove that the function $f(x) = \sin x + \cos x$ is a continuous function.

Let $g(x) = \sin x$ and $h(x) = \cos x$. We know that both $g(x)$ and $h(x)$ are continuous functions for all $x \in \mathbb{R}$. According to the algebra of continuous functions, if $g$ and $h$ are continuous, then their sum $(g + h)$ is also continuous. Therefore, $f(x) = g(x) + h(x) = \sin x + \cos x$ is continuous for all $x \in \mathbb{R}$.

This solution uses the theorem that the sum of two continuous functions is itself continuous.

Discuss the continuity of $f(x) = \sin(x^2)$.

Let $g(x) = \sin x$ and $h(x) = x^2$. Here, $h(x) = x^2$ is a polynomial function, so it is continuous for all $x \in \mathbb{R}$. Also, $g(x) = \sin x$ is a trigonometric function which is continuous for all $x \in \mathbb{R}$. The function $f(x)$ can be written as the composition of $g$ and $h$, i.e., $f(x) = (g \circ h)(x) = g(h(x)) = \sin(x^2)$. Since the composition of two continuous functions is continuous, $f(x) = \sin(x^2)$ is continuous for all $x \in \mathbb{R}$.

The continuity of the composite function is established by identifying the inner function (polynomial) and outer function (sine), both of which are known to be continuous.

Is the function $f(x) = \frac{x^2 + 2x + 1}{x - 1}$ continuous at $x = 1$?

Let $p(x) = x^2 + 2x + 1$ and $q(x) = x - 1$. Both $p(x)$ and $q(x)$ are polynomial functions and are continuous for all $x \in \mathbb{R}$. By the algebra of continuous functions, $f(x) = \frac{p(x)}{q(x)}$ is continuous at all points where $q(x) \neq 0$. At $x = 1$, $q(1) = 1 - 1 = 0$. Since the denominator becomes zero, the function $f(x)$ is not defined at $x = 1$. Therefore, $f(x)$ is not continuous at $x = 1$.

A rational function is only continuous at points where the denominator is non-zero. Since $x=1$ makes the denominator zero, the function is discontinuous (specifically, it has a point of discontinuity) there.

1. Let $g(x) = x - 2$, which is a polynomial function and hence continuous for all $x \in \mathbb{R}$.
2. Let $h(x) = |x|$, which is a known continuous function for all $x \in \mathbb{R}$.
3. The given function $f(x) = |x - 2|$ can be written as the composition $(h \circ g)(x) = h(g(x)) = |x - 2|$.
4. Since both $g$ and $h$ are continuous functions on $\mathbb{R}$, their composition $f = h \circ g$ is also continuous for all real numbers.

We use the property that the composition of two continuous functions is continuous. The absolute value of a polynomial is always continuous because the polynomial and the absolute value function are individually continuous.

1. We know that $\tan x = \frac{\sin x}{\cos x}$.
2. The functions $g(x) = \sin x$ and $h(x) = \cos x$ are continuous for all $x \in \mathbb{R}$.
3. According to the algebra of continuous functions, the quotient $\frac{g}{h}$ is continuous at all points where $h(x) \neq 0$.
4. $\cos x = 0$ when $x = (2n + 1)\frac{\pi}{2}$ for $n \in \mathbb{Z}$.
5. Therefore, $f(x) = \tan x$ is continuous at all points in its domain, which is $\mathbb{R} - \{(2n + 1)\frac{\pi}{2} : n \in \mathbb{Z}\}$.

Continuity of a quotient depends on the continuity of the numerator and denominator, as well as the denominator being non-zero. $\tan x$ is continuous wherever it is defined.