Unit 12.3 Calculus Topic 4: Increasing and decreasing functions: turning points · 2013. 11....

Unit 12.3 CalculusTopic 4: Increasing and decreasing functions:

turning pointsIn topic 4 we continue with straightforward derivatives and integrals:• Locate turning points where f’ (x) = 0.• Determine the nature of turning points.

Increasing and decreasing functionsA function y = f(x) is increasing on a set of values if, as the x-values increase, the function’s y-values increase.

A function y = f(x) is decreasing on a set of values if, as the x-values increase, the function’s y-values decrease.

Example AThe graph of y = f (x) is shown. At the points A (where x = 1) and B (where x = 8), the curve is neither increasing nor decreasing. The values x = 1 and x = 8 are called stationary values of the function.

y

x

y = f(x)

–4 –3 –2 –1 1 2 3 4 5 6 7 8 9 10 11

A

B

As the x-values increasefrom –4 to 1, the

y -values increase from1 to 5. The functiony = f (x) is said to be

increasing on theinterval –4 < x < 1.

As the x-values increase from1 to 8, the y -values decrease from

5 to 1. The function y = f(x) issaid to be decreasing on the

interval 1 < x < 8.

As the x-valuesincrease from 8,the functiony = f(x) is againan increasingfunction.

6

5

4

3

2

1

Turning pointsA point at which a function changes from increasing to decreasing or vice versa is called a turning point (abbreviation TP). Turning points are stationary values of the function.

If the function changes from increasing to decreasing, the turning point is called a local maximum. In Example A, the point A(1, 5) is a local maximum because the y-value at x = 1 is the maximum value of the function for all x-values close to 1.

Note: When the x-value exceeds about 11, the value of the function becomes greater than 5.

Local minimums are turning points where the function changes from being a decreasing function to being an increasing function. In Example A, the point B (8, 1) is a local minimum.

26_SAV_AM12_78720_TXT_4pp.indd 177 01/07/13 5:48 PM

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178 Unit 12.3 Calculus

© Oxford University Press www.oup.com.au

Tangents to graphsIn the diagram below, several tangents to the curve y = g(x) are drawn.

y

x

y = g (x)

–2 –1 1 2 3 4

function increasingfor x < 1; tangents

have positivegradients.

2

1

function decreasingfor 1 < x < 3;tangents have

negative gradients.

functionincreasing forx > 3; tangentshave positivegradients.

Any tangent drawn at a point on the graph of a function which is increasing is ‘tilted up’ and hence has a positive gradient. Thus at points on the graph of a function which is increasing, the derivative will be positive. Similarly when the function is decreasing the derivative is negative. This can be written:

If y = f (x) is an increasing function then f'(x) > 0.

Similarly, if y = f (x) is decreasing then f'(x) < 0.

At turning points, tangents are horizontal

x

yturningpoints

and have a gradient of 0.

At turning points, f'(x) = 0.

Determining the nature of turning pointsA graph will show clearly whether a turning point is a local maximum or minimum.

Example BQ. Find the turning point(s) of y = 2x2 – 8x + 4 and determine their nature.

A. y' = 4x – 8 [differentiating y = 2x2 – 8x + 4]

At TP, 4x – 8 = 0 [y' = 0 at the turning point(s)]

∴ x = 2 [solving for x]

At x = 2, y = 2 × 22 – 8 × 2 + 4 [substituting x = 2 in y = 2x2 – 8x + 4]

∴ y = –4

∴ coordinates of the TP are (2, – 4).

y = 2x2 – 8x + 4 is a quadratic

y

x

8

6

4

2

–2

–4

–1 1 2 3 4 5

function and its graph is a parabola.

The y-intercept is (0, 4) and from the diagram, (2, –4) is a local minimum.


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179Topic 4: Increasing and decreasing functions: turning points


Example CQ. 1. Find the turning points of y = 1

3x3 – x2 – 8x + 3 and determine their nature.

2. For what set of values is this function decreasing?

A. y' = x2 – 2x – 8 [differentiating y]

y' = 0 at any TP

∴ x2 – 2x – 8 = 0

∴ (x – 4)(x + 2) = 0 [factorising]

x = –2 or 4

when x = –2, y = 13(–2)3 – (–2)2 – 8(–2) + 3 [substituting x = –2 in y]

= –8

3 – 4 + 16 + 3

= 1213 ∴ TP is (–2, 121

3)

when x = 4, y = 13 × 43 – 42 – 8 × 4 + 3 [substituting x = 4 in y]

= –2323

∴ TP is (4, –2323

)

Substituting x = 0 in

y = 13

x3 – x2 – 8x + 3

shows that the function has a y-intercept of (0, 3).

A sketch of the graph

shows that (–2, 1213

)

is a local maximum

and (4, –2323

) is a

local minimum.

2. The function is decreasing for all values in the interval –2 < x < 4.

Another technique for determining the nature of turning points involves evaluating the gradient function in the neighbourhood of the turning point.

Example DQ. Find the turning points of y =

13

x3 – x2 – 8x + 3 and determine their nature.

A. From Example C, the turning points occur at x = 2 and x = 4. The table shows the values of the gradient function in the neighbourhoods of these two x-values.

x –3 –2 –1 3 4 5[for example f' (–3) = (–3)2 – 2 × (–3) – 8 = 7 (positive)]

f’ (x) + 0 – – 0 +

Between x = –3 andx = –1, the graph Between x = 3 and x = 5, the graph

changes from increasing (positive changes from decreasing (negative

gradient) to decreasing (negative gradient) to increasing (positive

gradient), so x = –2 is a maximum. gradient), so x = 4 is a minimum.

y

x

y = 13x3 – x2 – 8x + 3

40

30

20

10

–10

–20

–30

–40

–6 –5 –4 –3 –2 –1 1 3 4 5 6 7 8

(–2, 1213

)

(4, –2323

)

2


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So (–2, 1213) is a maximum, and (4, –23

23

) is a minimum turning point for the curve

y = 13x3 – x2 – 8x + 3.

Note: The nature of the turning points can also be determined by examining the values of the function in the neighbourhood of the turning points. For example, the table below shows that there is a minimum value at x = 4.

x 3 4 5

f (x) –21 –2323 –20

13

Unit 12.3 Activity 4A: Turning pointsFor each of the following functions, find the coordinates of the turning points and determine their nature. State the values for which the functions are increasing and those for which the functions are decreasing.1. y = x2 – 4x + 1 2. y = 6 – 8x – x2 3. y = 5 + 3x – 2x2

4. y = 4 – 3x – 2x2 – 13

x3 5. y = 8 – 6x – 52

x2 – 13

x3 6. y = 2x3 + 12x2 + 18x – 1

7. y = x3 – 12x + 2 8. y = 27x – x3 + 2 9. y = –3x3 – 3x2

10. a. State whether the curve y = 5 – 4x2 – x3 is increasing, decreasing or stationary when:

i. x = –223

ii. x = 2 iii. x = –112

iv. x = 0 b. Find the nature of any turning points.

11. f is a real number function such that f(x) = (x + 1)(x – 1)2 = x3 – x2 – x + 1. a. Sketch a graph of f, marking the intercepts on the axes. b. Differentiate to find f'(x). c. Find the values of x that give turning points of f, stating the nature of these

turning points. d. For what set of values of x is f a decreasing function? e. Find f(0). f. Find f'(0). g. Find the equation of the tangent to the graph of f, at x = 0.

12. y = Ax2 + Bx + C is a quadratic function. Its derivative, y', has a value of –2 when x = 1 and a value of 2 when x = 2. In addition, the point (1, –1) lies on the graph of this function. Find A, B and C.


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Alternative tests for turning pointsThe following are alternative tests for distinguishing maximum and minimum points of a graph and stationary points of inflection. In particular, these tests should be applied to the point x = a if it is found that f'(a) = f'(a) = 0 (ie test failure occurs). The sign of the gradient is considered in the neighbourhood of the stationary point.

Test for a maximum pointIf A is a maximum point at x = a, then the gradient of the graph changes from positive (+) to negative (–) as x increases, as shown in the figure. The test for a maximum point is made by using a numerical value for x just less than a and one for x just greater than a, and substituting each value into the derivative f'(x).

These values for the gradients are entered in the table alongside to verify the maximum point at x = a.

Test for a minimum pointIf B is a minimum point at x = b, then the gradient of the graph changes from negative (–) to positive (+) as x increases, as shown in the figure. The test for a minimum point is made using a numerical value for x just less than b, and one for x just greater than b, and substituting each value into the derivative f'(x).

These values for the gradients are entered in the table alongside to verify the minimum point at x = b.

Test for a stationary point of inflectionIf C is a stationary point of inflection at x = c, then the gradient of the graph stays positive (or negative) on both sides of the stationary point, as x increases.

0 x

y

++ + A

ax < a x > a

––

–

Sign of gradient

x x < a

f '(x) f '(x) > 0 f '(a) = 0 f '(x) < 0

Slope maximum

x > ax = a

0 x

y

bx < b x > b

–– – B +

++

Sign of gradient

x

f '(x) f '(x) < 0 f '(b) = 0 f '(x) > 0

Slope minimum

x < b x > bx = b

stationary point of in�ection

x

f '(x) f '(x) > 0 f '(c) = 0 f '(x) > 0

Slope

x < c x > cx = c

stationary point of in�ection

x

f '(x) f '(x) < 0 f '(c) = 0 f '(x) < 0

Slope

x < c x > cx = c

0 x

y

cx < c x > c

C

Sign of gradient (–)

–––

–––0

0 x

y

cx < c x > c

C

Sign of gradient (+)

+ + +

+++ 0


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PolynomialsA polynomial graph will have equation y = a

nxn + a

n – 1xn – 1 + ... + a

3x3 + a

2x2 + a

1x + a

0.

It is useful to have some idea of the graphs of polynomials. The following table shows examples that could represent some polynomials.

Unit 12.3 Activity 4B: Curve sketching1. Sketch a continuous curve y = f (x) with these properties: a. f(3) = 2, f'(3) = 0, f(x) is concave down for all x. b. f'(3) = f'(5) = 0, f(5) < 0 and f(3) = 0. c. f (3) = f'(3) = f''(3) = 0, f(x) is concave up for x > 3 and f(x) is concave down for

x < 3.

2. Find the coordinates of the turning points of f (x) = 5 + 12x – x3.

3. f(x) = x4 – 8x2 + 16 has three turning points. Find their coordinates and identify which are maximum points and which are minimum points. Show full working.

4. The equation of a curve is y = x(x – 2)(x + 3). a. State the x-intercepts and the y-intercept. b. Using calculus, find the x-coordinates of the maximum and minimum points and

distinguish between them using suitable tests. c. Sketch the curve showing the main features.

5. y = 2x(x – 2)(x + 2) has two turning points. a. Find their coordinates and identify which is which, using suitable tests. b. Identify the x-intercepts and y-intercept. c. Sketch the curve showing its properties.

Degree Graph Description

2 (quadratic) At most, one turning point.No points of in�ection.

3 (cubic) At most, two turning points.One point of in�ection.

4 (quartic) At most, three turning points.Two points of in�ection.

y

x0

0 x

y

0 x

y


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6. The equation of a curve is f(x) = 2x(2 + x)(3 – x). a. Find the x- and y-intercepts of the graph and explain how you obtained your

answers. b. Find the coordinates of the maximum and minimum points. c. Sketch the curve.

7. Find and identify the stationary points of the function f(x) = 2x5 – 5x4 + 20.

8. Find the coordinates of the turning points for y = x(x – 4)4.

9. The figure shows the graph of f (x) = x2(3 – x). The dotted line AB is a tangent to the curve at B, and y = 0 is a tangent to the curve at the origin. Calculate the coordinates of A and B (do not do a scale drawing).

10. The equation of a polynomial curve of degree 4 is y = 3x4 – 4x3. a. Find the coordinates of all the stationary points and distinguish between them.

Justify your answers. b. Explain why the x-intercepts are 0 and .

c. Find the non-stationary point of inflection. Justify your answer. d. Sketch the curve showing the main features. e. Write down the values of x where the curve is concave down.

11. Find the coordinates of the stationary points for the function f (x) = x4 – 9x2 and show tests for distinguishing the maximum, minimum and points of inflection. Find all inflection points, the intercepts, and hence sketch the curve.

12. a. Find and identify the stationary points of the function f (x) = x5 – x. If the curve crosses the x-axis at x = a, x = b, x = c, find a, b and c. Hence, sketch the curve.

b. From the graph of f (x) = x5 – x find all the values of x for which: i. f (x) is decreasing. ii. f(x) is concave up.

13. Investigate the turning points for the curve y = 2x4 + px2. a. For what values of p does this curve have three turning points? b. If p = –1 find the turning points and sketch the curve.

f(x)

x0

AB

43


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Unit 12.3 Calculus Topic 4: Increasing and decreasing functions: turning points · 2013. 11....

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