Continuity and Differentiability

A function is continuous where its graph has no break, and differentiable where the graph also has a well-defined tangent slope. Both are defined by limits, and the hypotheses of the mean value theorems, the tests for maxima and minima and the fundamental theorem of calculus are stated in these terms. The tasks are to check continuity or differentiability of a piecewise or absolute-value function at a point, find the constants that make a function smooth, count the points where a function fails to be differentiable, or test the "continuous but not differentiable" examples.

$f$ is continuous at $a$ if all three hold:

Equivalently, for every $\varepsilon>0$ there exists $\delta>0$ such that for every $x$ in the domain, $|x - a| < \delta \implies |f(x) - f(a)| < \varepsilon$. Right-continuity at $a$ uses only $\lim_{x \to a^+}$, and left-continuity only $\lim_{x \to a^-}$. $f$ is continuous on an open interval if it is continuous at every point, and on $[a, b]$ if in addition it is right-continuous at $a$ and left-continuous at $b$.

Sums, differences, products and constant multiples of continuous functions are continuous; so is a quotient wherever the denominator is non-zero, and so is a composition $g \circ f$ when $f$ is continuous at $a$ and $g$ at $f(a)$. Polynomials, $\sin$, $\cos$, $e^x$ and $|x|$ are continuous everywhere; $\ln x$ and $\sqrt{x}$ on their domains; rational functions and $\tan x$ wherever defined. $\lfloor x \rfloor$ is discontinuous at every integer (it is right-continuous there).

A removable discontinuity is repaired by redefining $f(a)$ as the limit; the others cannot be repaired.

Intermediate value theorem (IVT). If $f$ is continuous on $[a, b]$ and $N$ lies between $f(a)$ and $f(b)$, then $f(c) = N$ for some $c \in [a, b]$. In particular, if $f(a)$ and $f(b)$ have opposite signs, $f$ has a root in $(a, b)$; this is the basis of the bisection method.

Extreme value theorem. If $f$ is continuous on the closed, bounded interval $[a, b]$, it attains a maximum and a minimum value there. Both hypotheses matter: $1/x$ on $(0, 1]$ is unbounded, and $x$ on $(0, 1)$ has no maximum.

$f$ is differentiable at $a$ if

exists (finite). Geometrically, $f'(a)$ is the slope of the tangent at $(a, f(a))$. The left and right derivatives use $h \to 0^-$ and $h \to 0^+$; $f$ is differentiable at $a$ if and only if both exist and are equal.

Differentiable implies continuous. If $f'(a)$ exists, then $f(a + h) - f(a) = h \cdot \frac{f(a + h) - f(a)}{h} \to 0 \cdot f'(a) = 0$. The converse fails, in three typical ways:

Linearity, product $(uv)' = u'v + uv'$, quotient $(u/v)' = (u'v - uv')/v^2$, and chain rule $(g \circ f)'(x) = g'(f(x))f'(x)$. For an inverse function, $(f^{-1})'(y) = 1/f'(x)$ where $y = f(x)$ and $f'(x) \ne 0$.

For differentiable real functions $u,v$ with $u(x)>0$, to differentiate $u(x)^{v(x)}$, differentiate $\ln y = v \ln u$: $y' = y\,(v' \ln u + v u'/u)$. For example $(x^x)' = x^x(\ln x + 1)$.

Derivative versus continuous derivative. A function can be differentiable everywhere while $f'$ is not continuous. $f(x) = x^2\sin(1/x)$ with $f(0) = 0$ has $f'(0) = \lim h\sin(1/h) = 0$, yet $f'(x) = 2x\sin(1/x) - \cos(1/x)$ for $x \ne 0$ has no limit at 0.

At a junction point $a$ of a piecewise definition:

For $|g(x)|$ with $g$ differentiable: at a simple root of $g$ (where $g(a) = 0$, $g'(a) \ne 0$) the graph has a corner; away from roots $|g|$ is differentiable. At a root where $g'(a) = 0$ as well, $|g|$ is differentiable at $a$ (for example $|x^3|$ or $|x|x$).

$f(x) = \dfrac{\sin kx}{x}$ for $x \ne 0$ and $f(0) = 3$. Since $\lim_{x \to 0} \sin(kx)/x = k$, $f$ is continuous at 0 exactly when $k = 3$.

$f(x) = ax + b$ for $x < 1$, $f(x) = x^2$ for $x \ge 1$. Continuity at 1: $a + b = 1$. Equal one-sided derivatives: $a = 2$. So $a = 2$, $b = -1$, and the line is the tangent $y = 2x - 1$ to $y = x^2$ at $x = 1$.

$f(x) = |x - 1| + |x - 2|$ has slopes $-2$ on $(-\infty, 1)$, $0$ on $(1, 2)$ and $2$ on $(2, \infty)$. It is continuous everywhere and fails to be differentiable at exactly 2 points, $x = 1$ and $x = 2$. Likewise $|x^2 - 3x + 2| = |(x - 1)(x - 2)|$ has corners at its two simple roots 1 and 2.

$f(x) = x|x|$: $f(x) = x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$. Both one-sided derivatives at 0 are 0, so $f'(x) = 2|x|$ for all $x$, which is continuous. But $f'$ has a corner at 0, so $f''(0)$ does not exist.

$p(x) = x^3 - x - 1$ has $p(1) = -1 < 0$ and $p(2) = 5 > 0$, so a root lies in $(1, 2)$. Bisection: $p(1.5) = 0.875 > 0$, so the root is in $(1, 1.5)$; $p(1.25) \approx -0.297 < 0$ narrows it to $(1.25, 1.5)$. (The root is about 1.3247.)

For $f(x) = x^2\sin(1/x)$, $f(0) = 0$: $\dfrac{f(h) - f(0)}{h} = h\sin(1/h)$, and $|h\sin(1/h)| \le |h| \to 0$, so $f'(0) = 0$. For $g(x) = x\sin(1/x)$, $g(0) = 0$, the quotient is $\sin(1/h)$, which has no limit: $g$ is continuous at 0 (squeeze) but not differentiable there.

$\frac{d}{dx}\sin(x^2) = 2x\cos(x^2)$. $\frac{d}{dx}e^{\sin x} = \cos x\, e^{\sin x}$. For $y = x^x$ ($x > 0$), $\ln y = x \ln x$ gives $y'/y = \ln x + 1$, so $y'(1) = 1$ and $y'(e) = 2e^e$.