Differentiability and continuity: the derivative at a point as a two-sided limit, why differentiable implies continuous but not the reverse, and corners, cusps and vertical tangents, with piecewise functions (new explicit content in the 2024 Mathematics Advanced syllabus)

The pieces meet: $-0 + 1 = 1$ and $e^0 = 1$, so $f(0) = 1$ and $f$ is continuous. The gradient from the left is $-1$. The gradient from the right is $\frac{d}{dx} e^x = e^x$, which is $1$ at $x = 0$. Since $-1 \neq 1$, $f$ is not differentiable at $x = 0$: the graph has a corner at $(0, 1)$.

Markers give one mark for verifying continuity, one for each one-sided gradient evaluated at $x = 0$ and one for the conclusion.

Left piece at $x = 2$: $4 + 1 = 5$. Right piece: $4(2) - 3 = 5$. The values match, so $f$ is continuous at $x = 2$. The left gradient is $2x = 4$ and the right gradient is $4$. Both equal $4$, so $f$ is differentiable at $x = 2$ with $f^\prime(2) = 4$.

Markers award one mark for each of the two conditions being checked and one for the correct conclusion.

Find the gradient on each side of $x = 2$. For $x > 2$, $f(x) = x - 2$ and the gradient is $+1$. For $x < 2$, $f(x) = -(x - 2)$ and the gradient is $-1$.

The one-sided gradients disagree ($-1 \neq +1$), so the limit of $\frac{f(2+h) - f(2)}{h}$ as $h \to 0$ does not exist. The graph has a sharp corner at $(2, 0)$ and no single tangent there.

Marker's note: one mark for the two one-sided gradients, one for the conclusion. Saying "there is a corner" without the gradients is incomplete.

Continuity first. $x^2$ at $x = 1$ gives $1$, and $2x - 1$ at $x = 1$ gives $1$. The two pieces meet, so $f$ is continuous at $x = 1$.

Compare gradients. For $x < 1$, $f^\prime(x) = 2x$, which is $2$ at $x = 1$. For $x > 1$, $f^\prime(x) = 2$. Both one-sided gradients equal $2$, so $f$ is differentiable at $x = 1$ with $f^\prime(1) = 2$.

Marker's note: one mark for checking the pieces join, one for equal gradients. A straight line $y = 2x - 1$ is the tangent to the parabola at $(1, 1)$, which is why the join is smooth.

Continuous. $f(1) = 1$ and $3(1) - 2 = 1$, so the pieces meet at $(1, 1)$.

Not differentiable. The gradient from the left is $2x = 2$ at $x = 1$; the gradient from the right is $3$. Since $2 \neq 3$, there is no single derivative at $x = 1$: the graph has a corner.

Marker's note: one mark for continuity, one for each one-sided gradient. Do not conclude from continuity alone.

Differentiability needs two conditions at $x = 1$.

1. Continuity: the values match, so $1 = a + b$.
2. Equal gradients: the left gradient is $2x = 2$ and the right gradient is $a$, so $a = 2$.

Substituting, $b = 1 - 2 = -1$. So $a = 2$ and $b = -1$, and the right-hand piece is $2x - 1$.

Marker's note: one mark for each condition and one for solving. Using only the gradient condition and forgetting continuity gives the wrong $b$.

(a) Piecewise form. $x^2 - 4 \geq 0$ when $|x| \geq 2$, so

(b) Gradients at $x = 2$. Just to the left, $f(x) = 4 - x^2$ with gradient $-2x = -4$. Just to the right, $f(x) = x^2 - 4$ with gradient $2x = 4$.

(c) Points of non-differentiability. At $x = 2$ the one-sided gradients are $-4$ and $4$, which differ. At $x = -2$: just to the left $f(x) = x^2 - 4$ with gradient $2x = -4$, and just to the right $f(x) = 4 - x^2$ with gradient $-2x = 4$. These also differ, so $f$ has corners at $x = -2$ and $x = 2$ and is not differentiable at either.

At $x = 0$, $f^\prime(x) = -2x = 0$ and the value is $f(0) = 4$. Values nearby, such as $f(\pm 1) = 3$, are smaller, so $(0, 4)$ is a local maximum turning point.

Marker's note: one mark for the piecewise form, one for the one-sided gradients, one for naming both corners, one for the turning point.

(a) Use the definition. $f(0) = 0$, so

Both one-sided limits equal $0$, so $f$ is differentiable at $0$ with $f^\prime(0) = 0$. (Away from $0$, $f^\prime(x) = 2|x|$, which tends to $0$.)

(b) Continuity. $x^{1/3}$ is defined for all real $x$ and $g(h) \to 0 = g(0)$ as $h \to 0$, so it is continuous at $0$.

Non-differentiability. $\dfrac{g(h) - g(0)}{h} = \dfrac{h^{1/3}}{h} = h^{-2/3} = \dfrac{1}{h^{2/3}}$, which grows without bound as $h \to 0$ (from either side). The limit does not exist as a finite number; the tangent at the origin is vertical.

Marker's note: one mark for each limit setup and one for each conclusion. A vertical tangent means the gradient is undefined, so the function is not differentiable.