Maths › Differentiation › Tangents, turning points and curve behaviour
Tangents, turning points and curve behaviour
The derivative now gets used. It gives tangent and normal equations, locates the stationary points where curves turn, distinguishes maxima from minima by a second derivative, and finds the largest volume obtainable from a sheet of card, which is the kind of question these methods were developed for.
Builds on Differentiating powers of x and Straight lines.
IN THIS TOPIC
- Find equations of tangents and normals at a point on a curve.
- Locate stationary points and classify them with the second derivative.
- State the intervals on which a curve is increasing or decreasing.
- Use convex, concave and point of inflection correctly, and know what f'' = 0 does and does not prove.
- Solve practical maximisation and minimisation problems, domain included.
COMMON MISCONCEPTION
f''(x) = 0 means x is a point of inflection.
Tangents and normals
Evaluate the derivative at a point and you have a gradient. From there the straight-lines lesson does everything. The tangent at (a, f(a)) has gradient f'(a). The normal is perpendicular to it, so when f'(a) is non-zero its gradient is the negative reciprocal −1/f'(a), and both lines pass through the point itself. At a stationary point the tangent is horizontal and the reciprocal breaks down: the normal there is the vertical line x = a.
WORKED EXAMPLE
Both lines at one point
Find the equations of the tangent and the normal to y = x2 − 4x at the point where x = 3.
The point is (3, −3), and dy/dx = 2x − 4 gives gradient 2 there.
Tangent: y + 3 = 2(x − 3), so y = 2x − 9.
Normal: gradient −½, so y + 3 = −½(x − 3), which is y = −x/2 − 3/2.
Point-gradient form did the rest. The only calculus in the whole question was one evaluation of the derivative.
Stationary points, classified
Where f'(x) = 0 the curve is momentarily flat and you have a stationary point. The second derivative usually settles which kind it is. If f'' > 0 the gradient is increasing through zero, so the point is a minimum; if f'' < 0 it is a maximum. Both need the y-coordinate as well as the x, because a stationary point is a point.
GUIDED PRACTICE
Find and classify
Find the stationary points of y = x3 − 3x2 + 1 and classify each, before opening the working.
Show the working
dy/dx = 3x2 − 6x = 3x(x − 2), zero at x = 0 and x = 2, giving the points (0, 1) and (2, −3).
The second derivative is 6x − 6. At x = 0 it is −6, so (0, 1) is a maximum; at x = 2 it is +6, so (2, −3) is a minimum.
The curve also inflects at x = 1, where 6x − 6 changes sign, matching the figure point for point.
Increasing, decreasing, convex, concave
Over an interval, f' > 0 means the curve is increasing and f' < 0 means decreasing. One derivative up, f'' > 0 means convex, bending upwards like the inside of a bowl, and f'' < 0 means concave. A point of inflection is where the bending swaps over.
The f'' = 0 test needs two careful words. Where f'' exists at an inflection it is zero, and it can also fail to exist there, as it does at the origin of y = ∛x. But f'' = 0 on its own proves nothing whatever: y = x4 has f''(0) = 0 at what is plainly a minimum. The reliable test is whether f'' changes sign across the point. Examiners write the mark scheme around that phrase, so use it, and when you are stuck, check the gradient either side instead.
Optimisation, calculus with a purpose
Practical maximum and minimum questions all wear the same shape. Write the quantity you want to optimise as a function of one variable, differentiate, solve for zero, classify, and answer in context with units. Then look at the number and ask whether a box could really be that shape.
INDEPENDENT PRACTICE
The open box
Squares of side x cm are cut from the corners of a 10 cm square of card, and the sides fold up into an open box. Show that the volume is V = x(10 − 2x)2, and find the value of x that makes V greatest.
Show the working
The base is a (10 − 2x) square and the height is x, so V = x(10 − 2x)2, valid for 0 < x < 5.
Expanding and differentiating, or factorising directly, dV/dx = (10 − 2x)(10 − 6x), zero at x = 5 and x = 5/3. Only the second lies inside the domain, since x = 5 leaves no base at all.
d2V/dx2 at x = 5/3 is negative, so the volume is greatest at x = 5/3 cm, where V = 2000/27 ≈ 74.1 cm3.
State the domain and reject x = 5 with a reason. Optimisation marks are lost to context far more often than to calculus.
ASSESSMENT FOCUS
- Tangent gradient is f'(a), normal gradient is its negative reciprocal when f'(a) ≠ 0, and both lines then come out of point-gradient form. A horizontal tangent has the vertical normal x = a.
- Solve f'(x) = 0 fully and find the y-coordinates too. A stationary point is a point, never a bare x-value.
- Classify with the sign of f'' and say so out loud. “f''(2) = 6 > 0, so minimum” is the wording the mark scheme prints.
- f'' = 0 decides nothing by itself. Test for a sign change, or fall back on the gradient either side.
- Increasing and decreasing want an interval in the answer, written as an inequality in x, not a single value.
- In practical problems state the domain, reject stationary points outside it with a reason, and give the answer in context units.
CHECK YOURSELF
The curve y = x3 − 12x has two stationary points. Find and classify them, and state the interval on which the curve is decreasing.
Show a hint
Factorise the derivative, then read signs.
Show the answer
dy/dx = 3x2 − 12 = 3(x − 2)(x + 2), zero at x = ±2, giving (−2, 16) and (2, −16).
f'' = 6x, negative at −2, so (−2, 16) is a maximum; positive at 2, so (2, −16) is a minimum.
Between the roots the derivative is negative, so the curve is decreasing on −2 < x < 2.
Tangents take f'(a); normals take its negative reciprocal when f'(a) ≠ 0, and the vertical x = a when f'(a) = 0; flat points solve f'(x) = 0.
The second derivative's sign classifies; only a sign change makes an inflection.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the tangents, turning points and curve behaviour questions page.
CHECK YOUR PROGRESS
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- Find equations of tangents and normals at a point on a curve.
- Locate stationary points and classify them with the second derivative.
- State the intervals on which a curve is increasing or decreasing.
- Use convex, concave and point of inflection correctly, and know what f'' = 0 does and does not prove.
- Solve practical maximisation and minimisation problems, domain included.
Open the full revision checklist to see every objective in the course in one place.