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Tangents, turning points and curve behaviour

Now the derivative starts earning. It writes tangent and normal equations, finds the flat points where curves turn, tells hilltops from valleys with a second glance, and squeezes the most volume out of a sheet of card, which is the kind of question this machinery was invented for.

Year 12-13EDEXCEL 9MA0 7.3, 7.1

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.
  • Use increasing, decreasing, convex and concave language, and solve practical max/min problems.

WHAT YOU PROBABLY THINK

f''(x) = 0 means x is a point of inflection.

Tangents and normals

The derivative evaluated at a point is a gradient, and the straight-lines lesson does the rest. The tangent at (a, f(a)) has gradient f'(a); the normal, perpendicular to it, has the negative reciprocal gradient, and both lines pass through the point itself.

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 runs the whole question; the only calculus was one evaluation of the derivative.

Stationary points, classified

Where f'(x) = 0 the curve is momentarily flat, a stationary point, and the second derivative usually settles which kind. f'' > 0 means a minimum, the gradient is increasing through zero; f'' < 0 means a maximum. On intervals, f' > 0 means increasing and f'' > 0 means convex (bending upward), with concave the mirror case, and a point of inflection is where convexity changes.

The cubic y equals x cubed minus 3 x squared plus 1 with its maximum at 0 comma 1, its minimum at 2 comma minus 3, and its point of inflection at 1 comma minus 1 where the second derivative changes signmaximum (0, 1)minimum (2, −3)inflection (1, −1)zero gradient at both; the second derivative decides
FIG. 1The cubic y = x³ − 3x² + 1: maximum at (0, 1) where f'' < 0, minimum at (2, −3) where f'' > 0, inflection at (1, −1) where f'' changes sign.

The opening lie needs one careful word. At an inflection with a horizontal tangent f'' is indeed zero, but f'' = 0 alone proves nothing: y = x4 has f''(0) = 0 at what is plainly a minimum. The honest test is whether f'' changes sign, and when in doubt, examine the gradient either side.

YOUR TURN

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, negative, 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.

Optimisation: calculus with a purpose

Practical maximum and minimum questions all share one shape: write the quantity to optimise as a function of a single variable, differentiate, set the derivative to zero, classify, and answer in context, checking the value makes physical sense.

The volume of an open box folded from a 10 by 10 sheet with corner squares of side x: V equals x times bracket 10 minus 2x squared, greatest at x equal to five thirds where the volume reaches 74.1greatest volume 74.1x = 5/3V = x(10 − 2x)²
FIG. 2An open box folded from a 10 cm square: volume x(10 − 2x)², greatest at x = 5/3, where the calculus and the curve agree on 74.1 cm³.

TRY IT UNSEEN

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 (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.

Stating the domain and discarding x = 5 with a reason are answer-standard here; optimisation marks are lost to context more than to calculus.

THE EXAM BIT

  • Tangent gradient is f'(a); normal gradient is its negative reciprocal; both lines then come from point-gradient form.
  • 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: “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.
  • 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 exactly on −2 < x < 2.

Tangents take f'(a); normals take its negative reciprocal; flat points solve f'(x) = 0.

The second derivative's sign classifies; only a sign change makes an inflection.

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.
  • Use increasing, decreasing, convex and concave language, and solve practical max/min problems.

No animated video for this topic yet; these notes stand alone.