Topics

Continuity, differentiability and concavity

Covers AS91578

Continuity in one sentence

A function is continuous at $x = a$ if you can draw through that point without lifting your pen.
Formally, all three of these must hold:
- $f(a)$ is defined
- $\displaystyle\lim_{x \to a} f(x)$ exists
- the two are equal: $\displaystyle\lim_{x \to a} f(x) = f(a)$
If any one fails, the function is discontinuous at $a$ .

Differentiability

A function is differentiable at $x = a$ if $f'(a)$ exists — that is, if the curve has a single well-defined tangent there.
The graph must be smooth, with no corner, cusp or break.

The one-way relationship

Differentiable $\Rightarrow$ continuous. If a curve has a tangent at a point, it cannot have a break there.
Continuous $\not\Rightarrow$ differentiable. A curve can be unbroken and still have a sharp corner.
The standard counterexample is $y = |x|$ at $x = 0$ : unbroken, but the gradient is $-1$ on the left and $+1$ on the right, so there is no single gradient at the origin.
The three ways differentiability fails while continuity holds:
- Corner — the left and right gradients exist but differ
- Cusp — the gradients head to $+\infty$ and $-\infty$
- Vertical tangent — the gradient is infinite, so $f'(a)$ does not exist as a number

Testing a piecewise function for differentiability

Check continuity first. If the pieces do not join, differentiability is impossible — stop there.
Differentiate each piece separately.
Evaluate both derivatives at the joining point.
Compare. Equal means differentiable; unequal means a corner.
Worked through for $f(x) = x^2$ for $x \leq 1$ and $f(x) = 2x - 1$ for $x > 1$ :
- Continuity: $1^2 = 1$ and $2(1) - 1 = 1$ ✓ they join
- Gradients: $2x$ gives 2 at $x = 1$ ; the constant gradient of the line is 2
- Equal, so $f$ is differentiable at $x = 1$ — the line is exactly the tangent to the parabola there.

Concavity

Concavity describes which way a curve bends, and it is read from the second derivative $f''(x)$ .
Concave up ( $f''(x) > 0$ ) — the curve bends upwards, like a valley or a cup that holds water. The gradient is increasing.
Concave down ( $f''(x) < 0$ ) — the curve bends downwards, like a hill. The gradient is decreasing.
Note the distinction that catches people out: concavity is about whether the gradient is increasing, not whether the function is increasing. A curve can be falling steeply and still be concave up, if it is falling less and less steeply.

What the two derivatives tell you

	$f'(x) > 0$	$f'(x) < 0$
$f''(x) > 0$	increasing, concave up	decreasing, concave up
$f''(x) < 0$	increasing, concave down	decreasing, concave down

$f'$ answers "is it going up or down?"
$f''$ answers "which way is it bending?"

Increasing and decreasing intervals

A function is increasing on an interval where $f'(x) > 0$ , and decreasing where $f'(x) < 0$ .
To find the intervals:
- Differentiate and set $f'(x) = 0$ to find the boundaries.
- Test the sign of $f'$ in each interval between them.
- State the intervals in terms of $x$ .
The 2025 report lists "provided the conditions, using the derivative, for when a function is decreasing" as an Achieved-level behaviour, so these questions appear regularly.

Reading properties off a graph

Exam questions frequently supply a piecewise graph and ask you to read features from it. What to look for:
- Break in the curve — discontinuous, and therefore not differentiable
- Sharp corner — continuous but not differentiable
- Horizontal tangent — $f'(x) = 0$ , a stationary point
- Curve bending upwards — $f'' > 0$
- Change in bending direction — a point of inflection

Tips and Tricks

Continuity and differentiability are different questions, in that order. Check continuity first; if the pieces do not meet, you are done. The report notes candidates who "did not recognise features of gradient, continuity, and limits from a piece-wise graph" — the fix is to work through the checks in a fixed order rather than eyeballing the picture.

A corner is the trap. The function is perfectly continuous there, so a candidate who only checks continuity will wrongly conclude it is differentiable. Always compare the two one-sided gradients.

Concave up does not mean increasing. Sketch $y = e^{-x}$ if you need convincing: it is decreasing everywhere and concave up everywhere.

When asked where a function is increasing, give an interval in $x$ , such as $x < -2$ or $1 < x < 4$ — not a list of points, and not a statement about $y$ .

If $f''(a) = 0$ , you have not automatically found an inflection point. The concavity must actually change sign there. $y = x^4$ has $f''(0) = 0$ but is concave up on both sides, so the origin is a minimum, not an inflection.

Worked ExampleTesting a piecewise function

The function $f$ is defined by $f(x) = \begin{cases} x^2 + 2 & x \leq 1 \\ 4x - 1 & x > 1 \end{cases}$ Determine whether $f$ is (a) continuous and (b) differentiable at $x = 1$ .

Step 1 — Test continuity: the left-hand limit

For $x \leq 1$ the first piece applies:

$\lim_{x \to 1^-} f(x) = 1^2 + 2 = 3$

Step 2 — Test continuity: the right-hand limit

For $x > 1$ the second piece applies:

$\lim_{x \to 1^+} f(x) = 4(1) - 1 = 3$

Step 3 — Compare, and check the function value

Both one-sided limits are 3, so the two-sided limit exists and equals 3.

The function value at $x = 1$ comes from the first piece (because it is defined for $x \leq 1$ ):

$f(1) = 1 + 2 = 3$

All three agree, so $f$ is continuous at $x = 1$ .

Step 4 — Differentiate each piece

$\frac{d}{dx}(x^2 + 2) = 2x \qquad \frac{d}{dx}(4x - 1) = 4$

Step 5 — Evaluate both at the joining point

Left-hand gradient, from the parabola:

$f'(1^-) = 2(1) = 2$

Right-hand gradient, from the line — a line has the same gradient everywhere:

$f'(1^+) = 4$

Step 6 — Compare

$2 \neq 4$

The one-sided gradients are different, so there is no single tangent at $x = 1$ and $f$ is not differentiable there.

Step 7 — Interpret

The graph has a corner at $(1, 3)$ : the parabola arrives with gradient 2 and the line leaves with gradient 4, so the curve changes direction abruptly without breaking.