How do I find an oblique asymptote?

Divide the numerator by the denominator. If the quotient is mx + c and the remainder term tends to 0 as x tends to infinity, then y = mx + c is the oblique asymptote.

What must a DSE M2 curve sketch show?

Whatever the question asks for, typically the intercepts, maximum and minimum points, points of inflexion and asymptotes. The curriculum lists symmetry and restrictions on x and y as further points to consider, but not every sketch needs all of them.

When is the second derivative test not enough?

When f″ is 0 at the stationary point. Then check the sign of f′ just before and just after the point: positive to negative is a maximum, negative to positive is a minimum.

Applications of Differentiation (DSE M2): Tangents, Maxima and Minima, Curve Sketching

Set f′(x) = 0 to find stationary points, then classify them: f″ > 0 is a minimum, f″ < 0 a maximum, and if f″ = 0 use the sign of f′ on each side. For curve sketching also find the intercepts, points of inflexion (where f″ changes sign) and the vertical, horizontal and oblique asymptotes.

Solve Applications of Differentiation questions with AI

y = (x² + 3)/(x − 1) from Example 1, with its asymptotes x = 1 and y = x + 1, the maximum (−1, −2) and the minimum (3, 6). Try it yourself →

Applications of Differentiation: Key Facts

Tangent at x = x₀y − f(x₀) = f′(x₀)(x − x₀)

Stationary pointf′(x) = 0

Second derivative testf″ > 0: minimum; f″ < 0: maximum; f″ = 0: no conclusion

Point of inflexionf″ changes sign (concavity changes)

Vertical asymptotex = a where the denominator is 0 and the numerator is not

Oblique asymptoteLong division: f(x) = mx + c + (remainder)/(divisor) gives y = mx + c

Rate of changedV/dt = (dV/dr)(dr/dt)

Worked Examples — Applications of Differentiation

Example 1: Sketching a rational function with an oblique asymptote

Let y = (x² + 3) / (x − 1). Find the asymptotes and the maximum and minimum points of the curve.

Long division: y = x + 1 + 4/(x − 1), so the asymptotes are x = 1 and y = x + 1
dy/dx = 1 − 4/(x − 1)²
dy/dx = 0 gives (x − 1)² = 4, so x = 3 or x = −1
d²y/dx² = 8/(x − 1)³: positive at x = 3 (minimum), negative at x = −1 (maximum)
y(3) = 6, y(−1) = −2

Answer: Asymptotes x = 1, y = x + 1; minimum point (3, 6), maximum point (−1, −2)

Example 2: A rate of change problem

Air is pumped into a spherical balloon at 100 cm³/s. Find the rate at which the radius is increasing when the radius is 5 cm.

V = (4/3)πr³, dV/dr = 4πr²
dV/dt = 4πr² · dr/dt
100 = 4π(5)² · dr/dt
dr/dt = 100 / (100π) = 1/π

Answer: 1/π cm/s (≈ 0.318 cm/s)

💡 Exam Traps & Secrets

⚠️ A local maximum can be below a local minimumIn Example 1 the maximum value is −2 and the minimum value is 6. On a curve with a vertical asymptote the two branches are separate, so "maximum" and "minimum" are local, not global.

⚠️ f″ = 0 does not guarantee a point of inflexionYou must show that f″ changes sign. For y = x⁴, f″(0) = 0 but the curve is concave upwards on both sides, so (0, 0) is a minimum, not a point of inflexion.

✏️ Self-Test: Quick Interactive Practice

How to study: Click any question below to instantly load it on the coordinate grid and check your steps.

(Curve sketching)Sketch y = x³ − 3x² + 4, showing turning points and the point of inflexion.

(Tangent from an external point)Find the equations of the tangents to y = x² + 1 that pass through (1, −2).

(Optimisation)A closed cylinder has volume 250π cm³. Find the radius that minimises its surface area.