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For f(x) = 2x³ − 3x² − 12x + 5, the derivative factorises as 6(x − 2)(x + 1). What are the critical points?
x = −1 and x = 2.
Use the first derivative test at x = −1. How does the sign of f′ change there?
It goes from positive to negative, so there is a local maximum at x = −1.
Now confirm with f″(x) = 12x − 6.
f″(−1) = −18, which is negative, so it is a maximum. And f″(2) = 18, so x = 2 is a minimum.
Good. Never stop at the point: work out the values too. f(−1) = 12 and f(2) = −15.
Take f(x) = x⁴. Where is f′(x) zero, and what is f″ there?
f′(x) = 4x³ is zero at x = 0, and f″(0) = 0. So the test fails.
Then what do we do?
Go back to the first derivative test. f′ is negative before 0 and positive after, so x = 0 is a local minimum.
Exactly. Now try x³. Its derivative 3x² is zero at 0 as well.
But 3x² is positive on both sides, so the sign never changes. There is no maximum or minimum, only a point of inflection.
We cut squares of side x from the corners of an 18 cm square sheet and fold up the sides. Write the volume in terms of x.
The base becomes 18 − 2x on each side and the height is x, so V = x(18 − 2x)².
What values of x make sense?
x must be between 0 and 9, otherwise there is no base left.
Differentiating gives V′(x) = 12(x − 3)(x − 9). Which critical point do we keep?
Only x = 3, since 9 is outside the range. V″(3) is −72, so it is a maximum, and the volume is 3 × 12², which is 432 cubic centimetres.
Perfect. Always end an optimisation problem by answering the question in words, with units.
a sign change of f′ from + to − at c gives a local maximum, from − to + a local minimum, and no change gives neither.
if f′(c) = 0, then f″(c) < 0 gives a maximum, f″(c) > 0 a minimum, and f″(c) = 0 means the test fails.
the extreme point is x = c, while the maximum or minimum value is f(c); give both in answers.
choose a variable, write the quantity in that one variable, state its range, differentiate, check, and answer in words.
Absolute extremes on [a, b]: compare f at the critical points inside the interval and at both end points.
f(x) = 2x³ − 3x² − 12x + 5 has f′(x) = 6(x − 2)(x + 1), a local maximum value of 12 at x = −1 and a local minimum value of −15 at x = 2.
f(x) = x⁴ has f″(0) = 0, so the second derivative test fails, but f′ changes from negative to positive at 0, giving a local minimum of 0.
An open box folded from an 18 cm square sheet has the greatest volume, 432 cm³, when squares of side 3 cm are cut from the corners.
On [1, 5], f(x) = 2x³ − 15x² + 36x + 1 has its absolute maximum 56 at x = 5 and its absolute minimum 24 at x = 1.
Plus then minus, a peak is found; minus then plus, the valley's ground.
A tin-smith had an 18 cm square sheet and wanted the roomiest box. Tiny corner cuts gave a flat tray, and big cuts gave a narrow tower. Calculus told him to cut exactly 3 cm, and his box held 432 cubic centimetres.
Frown for a crown, smile for a pile: a negative f″ (a frowning curve) gives a maximum, a positive f″ (a smiling curve) gives a minimum.
V.E.R.D.A. - Variable, Expression in one variable, Range, Differentiate, Answer in words.
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f′ changes from positive to negative at x = c. What does f have there?
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f′(c) = 0 and f″(c) > 0. What does f have at c?
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What should you do if f′(c) = 0 and f″(c) = 0?
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What is the largest open box that can be folded from an 18 cm square sheet?
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How do you find the absolute maximum of f on [a, b]?
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A point where a function's value is greater than at all nearby points.
A point where a function's value is less than at all nearby points.
The value f(c) of the function at a point of maximum c.
A point where a curve changes the way it bends, such as x = 0 on the graph of y = x³.
The greatest value a function takes on its whole domain or on a given interval.
Finding the largest or smallest possible value of a quantity, such as area, volume or cost.
A condition that limits the possible values in an optimisation problem, such as a fixed perimeter or sheet size.
Worked examples that find and classify local maxima and minima with the first and second derivative tests, and then use them to solve real optimisation problems.
This lesson puts the first and second derivative tests to work. It classifies the critical points of a cubic using both tests, shows what to do when the second derivative test fails (as for x⁴ and x³), finds an extreme value of a trigonometric function, and solves two optimisation problems: the largest open box that can be folded from a square sheet, and two numbers with a fixed sum whose squares add up to as little as possible. It ends by comparing local extremes with the absolute maximum and minimum of a function on a closed interval.
Let f(x) = 2x³ − 3x² − 12x + 5. Then f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1), so the critical points are x = −1 and x = 2.
The local maximum value is 12 at x = −1, and the local minimum value is −15 at x = 2.
For f(x) = x⁴, f′(x) = 4x³ = 0 only at x = 0, and f″(x) = 12x², so f″(0) = 0 and the second derivative test gives no answer. Using the first derivative test, f′(x) < 0 for x < 0 and f′(x) > 0 for x > 0, so x = 0 is a local minimum with value 0.
Compare f(x) = x³. Again f′(0) = 0 and f″(0) = 0, but f′(x) = 3x² is positive on both sides of 0. The sign does not change, so x = 0 is neither a maximum nor a minimum. It is a point of inflection, where the curve changes the way it bends.
For f(x) = sin x + cos x with 0 < x < π/2, f′(x) = cos x − sin x = 0 gives tan x = 1, so x = π/4. Then f″(x) = −sin x − cos x, and f″(π/4) = −√2 < 0, so there is a local maximum. Its value is f(π/4) = √2/2 + √2/2 = √2.
Equal squares are cut from the corners of a square sheet of tin of side 18 cm, and the sides are folded up to make an open box. What size of square gives the greatest volume?
Squares of side 3 cm should be cut off, giving the largest volume of 432 cm³.
Find two positive numbers whose sum is 15 and the sum of whose squares is least. If one number is x, the other is 15 − x, and S(x) = x² + (15 − x)². Then S′(x) = 2x − 2(15 − x) = 4x − 30 = 0 gives x = 7.5, and S″(x) = 4 > 0, so this is a minimum. The numbers are 7.5 and 7.5, and the least sum of squares is 112.5.
On a closed interval [a, b], the absolute maximum and minimum are found by comparing f at the critical points inside the interval and at the end points. For f(x) = 2x³ − 15x² + 36x + 1 on [1, 5], f′(x) = 6(x − 2)(x − 3). The values are f(1) = 24, f(2) = 29, f(3) = 28 and f(5) = 56. So f has a local maximum of 29 at x = 2 and a local minimum of 28 at x = 3, but its absolute maximum on [1, 5] is 56 at x = 5 and its absolute minimum is 24 at x = 1. A local maximum need not be the greatest value overall.