1. Finding Equation of the tangent to a curve
The gradient of a tangent to the
curve y = f (x)
is given by
f (x) or
for all values of x

2. dy/dx gives the gradient for all values of x.
Finding Equation of the tangent to a curve
A tangent to a curve is a straight line
The straight line equation is y – b = m(x – a)
To use the equation you need a point on the line (a , b) and the gradient, m
dy/dx gives the gradient for all values of x.

3. The gradient for all values of x
Find the gradient of the tangent to the curve y = x2 – 5x at each of the points A and B.
The gradient for all values of x
Gradient of tangent :
B(-1,6)
A(3,-6) A
When x = 3 ,
= 1
When x = -1, B
= -7
So gradient of tangent at A is 1 and gradient of tangent at B is -7

4. Find the equation of the tangent to the curve y = x2 – 5x at each of the points A and B
Tangent at A
B(-1,6)
A(3,-6)
Equation of line is
y - b = m(x - a)
m = 1 a b
Point on line is (3, -6)
y + 6 = 1(x – 3)
Tangent at B
y = x – 9
m = – 7
(a, b) = (– 1 , 6)
y – 6 = – 7(x +1)
y = – 7x – 1

5. Find the equation of the tangent to the curve y = x3 – 5x + 3 at the point where x = 1.
Gradient of tangent
To find the y-coordinate of the point on the curve, substitute x = 1 in y = x3 – 5x + 3
When x = 1,
When x = 1,
m = 3×1 – 5
y = 1 – 5 × 1 + 3 a b
y = – 1
( 1 , – 1)
m = – 2
Using y – b = m(x – a)
y + 1 = –2(x – 1)
y = –2x + 1)

6. Exercise 5A,
page 67
Questions
3 to 7
inclusive

7. Tangent at A m = 4 Equation is y = 4x + 1
1 Find the gradient of the tangent to the curve y = x2 + 5 at each of the points A and B. Find the equation of each tangent. y
y = x2 + 5
B (-4, 21) 
Answers
= 2x 
A (2, 9) x
Tangent at A m = 4 Equation is y = 4x + 1
Tangent at B m = –8 Equation is y = –8x – 11

8. But we are told the gradient is 9
Find the point of contact of the tangent to the curve y = x2 + 3x + 5 which has gradient 9.
But we are told the gradient is 9
y- coordinate is got from original equation
y = ×3 + 5
Point of contact is (3 , 23)
y = 23

9. But we are told the gradient is 10
Find the points of contact of the tangent to the curve y = x3 – 2x which has gradient 10.
But we are told the gradient is 10
y- coordinate is got from original equation
y = 23 – 2×2
y = (-2)3 – 2×(-2)
y = 4
y = –4
Points of contact are (2 , 4) or (-2 , -4)

10. Tangent at E( 2 , 5) m = -4 Equation is y = -4x + 13
2 Find the gradient of the tangent to the curve y = (3 + x)(3 - x) at each of the points E and F. Find the equation of each tangent. y
y = (3 + x)(3 - x)
Answers
E (2, 5) 
F (-3 , 0)  x
= -2x
Tangent at E( 2 , 5) m = -4 Equation is y = -4x + 13
Tangent at F (-3, 0) m = 6 Equation is y = 6x + 18

11. Tangent at A m = 0 Equation is y = 4
3 Find the gradient of the tangent to the curve y = x4 - 5x2 + 4 at each of the points A , B and C. Find the equation of each tangent. y
y = x4 – 5x2 + 4
A (0, 4) 
Answers
B (-2, 0)
C (1, 0)  
= 4x3 - 10x x
Tangent at A m = 0 Equation is y = 4
Tangent at B m = -12 Equation is y = -12x - 24
Tangent at C m = -6 Equation is y = -6x + 6

12. Increasing and Decreasing Functions
dy dx
f (x)

13. Increasing and decreasing functions
For any curve :
m is positive
m is negative
m is zero
Remember graphs are read from left to right!

14. Increasing and decreasing functions
f’(x) > 0, then f(x) is increasing as x increases
f’(x) < 0, then f(x) is decreasing as x increases
f’(x) = 0, then f(x) is stationary

15. f(x) is never decreasing since f ‘(x) ≥ 0
Show that f(x) = 1/3x3 – x2 + x – 1 is never decreasing?
f ‘ (x) = x2 – 2x + 1
f ‘ (x) = (x – 1)(x – 1)
f ‘ (x) = (x – 1)2
Can never be negative
f(x) is never decreasing since f ‘(x) ≥ 0

16. f (x) is never increasing
Show that f(x) = 1 – 6x + 12x2 – 8x3 is decreasing for all values of x except when x = ½.
f ’(x) = – x – 24x2
Squared term can never be negative
Consider the graph of y = – x – 24x2
f ’(x) = 0 when 2x – 1 = 0
ie x = 1/2
– x – 24x2 = 0
f (x) is decreasing for all values of x except x = ½, since f '(x) < 0 for all other values of x.
– 6 (1 – x + 4x2) = 0
– 6 (4x2 – x + 1) = 0
– 6 (2x –1) (2x –1) = 0
f (x) is never increasing
– 6 (2x –1)2 = 0

17. Consider the graph of y = – 6 + 24x – 24x2
Show that f(x) = 1 – 6x + 12x2 – 8x3 is decreasing for all values of x except when x = ½.
f ’(x) = – x – 24x2
f ’(x) = 0 when x = ½
Consider the graph of y = – x – 24x2
f (x) is decreasing for all values of x except x = ½, since f '(x) < 0 for all other values of x.
Graph of f '(x) is always below x-axis except when x = ½
f (x) is never increasing

18. f(x) decreasing for – 2 < x < 2
For what values of x is f(x) = x3 – 12x + 2 decreasing?
f ‘ (x) = 3x2 – 12
f(x) decreasing  f ‘ (x) < 0
Sketch f‘ (x) = 0
3x2 – 12 = 0
3(x2 – 4) = 0
3(x + 2)(x – 2) = 0
x = 2 or x = – 2
f(x) decreasing when f ‘ (x) < 0
f(x) decreasing for – 2 < x < 2

19. Stationary points
dy dx
f (x)

20. The Rule for Stationary Points Stationary points occur when
f (x) = 0 or = 0

21. This is a maximum turning point
To find the nature ( maximum, minimum, point of inflection) of a stationary point, we consider the gradient of the tangent before and after the stationary point. y
= 0 x
before
after  + –
before
after  
- ve
+ ve
slope x
This is a maximum turning point

22. This is a minimum turning point
The nature of a stationary point is found by considering the gradient of the tangent before and after the stationary point. y x
before
after – +
before
after
- ve  
+ ve
slope x
This is a minimum turning point 
= 0

23. This is a maximum turning point This is a minimum turning point
To determine the nature of a stationary point, find out the gradient of the tangent before and after the stationary point. x
before x
after
before
after + - - +
slope
slope
This is a maximum turning point
This is a minimum turning point

24. Find the stationary point on the curve with equation y = x2 - 6x + 5.
Example 1
Find the stationary point on the curve with equation y = x2 - 6x + 5.
Differentiate the function
When x = 3,
y = x2 – 6x + 5 = 2x
– 6
– 6(3)
y =
(3)2
+ 5
For stationary values,
= 0
= 9 –
= – 4
ie 2x - 6 = 0
x = 3
So a stationary point occurs at (3, -4).

25. Stationary point occurs at (3, -4).
Example 1 (continued)
To find the nature of the stationary point on the curve with equation y = x2 - 6x + 5, draw a shape table.
Stationary point occurs at (3, -4).
Shape → 3 x dx dy – +
(3, -4) is a MINIMUM   

26. Now find y-coordinates
Example 2
Find the stationary points on the curve with equation y = 1/3x3 - x2 - 3x + 10 and determine their nature
Now find y-coordinates
When x = 3
y = 1/333 – 32 – 3 ×
y = 1
S.P. (3 , 1)
When x = – 1
x2 – 2x – 3 = 0
y = 1/3(-1)3 – (-1)2 – 3 × (-1) + 10
(x – 3)(x + 1) = 0
y = 112/3
S.P. (– 1 , 112/3)
x = 3 or x = – 1

27. Example 2 cont’d
To find the nature of the S.P.’s of the equation y = 1/3x3 - x2 - 3x + 10
S.P. (– 1 , 112/3)
S.P. (3 , 1) x →
– 1 3
Shape – + +     
(-1, 112/3) is a MAXIMUM
(3, 1) is a MINIMUM

28. y = 1/3x3 - x2 - 3x + 10
(– 1 , 112/3 )
(3 , 1 )

29.      + + – x → 3 Shape 12x2 – 4x3 = 0 4x2( 3 – x) = 0
Find the stationary points on y = 4x3 – x4 and determine their nature x → 3
Shape + + –     
12x2 – 4x3 = 0
4x2( 3 – x) = 0
(3 , 27) is a MAX
(0, 0) is a POINT OF INFLECTION
x = 0 or x = 3
x = 0  y = 0
x = 3  y = 27

30. y = 4x3 – x4
A point of inflection occurs when the there is a horizontal tangent, S.P., and the curve continues in the same direction before and after the stationary point.

31. Sketching the curve of a given function
The steps you need to follow:
1. Find the y-intercept by making x = 0
2. Find the x-intercept(s) by making y = 0
3. Find the stationary points, SP’s, using differentiation
4. Determine the nature of the SP’s using shape table
5. Determine the behaviour of the curve for large x and y.
6. Annotate your sketch showing all the main features.

32. Sketch and annotate fully the curve
1. Find the y-intercept by making x = 0
2. Find the x-intercept(s) by making y = 0
 x = –1 (twice), x = 2 x y –1 2 –2

33. Sketch and annotate fully the curve
3. Find the stationary points, S.P.’s, using differentiation x y –1 2 –2
x = –1  y = 0
(1 , –4)
x = 1  y = –4

34.      + – + Sketch and annotate fully the curve
4. Determine the nature of the S.P.’s using shape table x → -1 1
Shape + – +    x y  
(-1, 0) is a MAX
(1 , -4) is a MIN –1 2 –2
(1 , –4)

35. Sketch and annotate fully the curve
5. Determine the behaviour of the curve for large x and y.
As x → ∞, y → x3
As x → +∞, y → +∞
As x → –∞, y → –∞
6. Annotate your sketch showing all the main features. x y
MAX –1 2 –2
(1 , –4)
MIN

36. Total Surface area = 2 circles + curved area
A company want to redesign their tins to reduce the cost.
The tins are to be cylindrical and hold 128 cm3 of juice. Given that the curved surface area of a cylinder is given by the expression C = 2πrh, show that the total surface area of the tin is given by
b) Find the dimensions which reduce the cost of metal to a minimum.
Total Surface area = circles + curved area

37. b) Find the dimensions which reduce the cost of metal to a minimum.
Maximum or Minimum means differentiate and find SP’s and their nature.
r ≠ 0, so multiply through by r2
÷ by 4π

38. b) Find the dimensions which reduce the cost of metal to a minimum.
We now need to show r = 4 gives a minimum
Shape → 4 r dr dA – +   
r = 4 MIN
Can has radius, r = 4 cm, height, h = 2∙55 cm

39. Show that the area of the red triangle is given by: C F x D6 E
Find the greatest and least areas of the triangle BEF x B 8 A
Area = Rectangle – 3 triangles

40. The largest area since all other x values  24 – something
SP’s  H’(x) = 0
Shape
H’(x) → 4 x – +   
x = 4 MIN
When x = 4, H(x) = 24 –
= 16
The largest area since all other x values  24 – something
When x = 0, H(x) = 24

41. The following
questions are
taken from
past papers

42. 1. For what values of x is the function f(x) = x3 – 12x + 2 decreasing?
2. For the function f(x) = 3x4 + 2x3
a) Find the stationary values and determine their nature.
b) Find the points where the curve cuts the x and y axis.
c) Sketch the curve.
3. a) Show that x-1 is a factor of x3 – 6x2 + 9x – 4 and find the other factors.
b) Find the points where f(x) cuts the coordinate axes.
c) Find the stationary values and determine their nature.
d) Sketch the curve.
4. A function f is defined by f(x) = (x + 3)2(x – 1).
a) Find the points where f(x) cuts the coordinate axes.
b) Find the stationary values and determine their nature.
c) Sketch the curve.

43. f(x) = 2x3 -3x2 – 36x is increasing.
5. The diagram shows part of the
graph of y = 2x2(x - 3).
Determine the stationary values
and their nature. y x y
6. Part of the graph of y = x4 – 2x3 + 2x - 1
is shown in the diagram. Determine the
stationary values and justify their nature. x
7. Find algebraically the values of x for which the function
f(x) = 2x3 -3x2 – 36x is increasing. y
8. The curve shown in the diagram has equation
y = -x4 + 4x3 – 2.
Find the stationary values and determine their nature. x

44. 9. A curve has equation y = 2x3 + 3x2 + 4x – 5.
Prove that it has no stationary values.

45. Sketching the derivative
Sketch the graph of the derivative of the function f(x) whose graph is shown.
f(x) is a quadratic
f ’(x) = 0 at SP y 4 x
(2,2)
To the left of SP curve is increasing, f ’(x) > 0
To the right of SP curve is decreasing, f ’(x) < 0
Now draw smooth curve through shaded areas and zero on x-axis
y = f ’(x)

46. At SP’s f ’(x) = 0 Inceasing  f ’(x) > 0 Deceasing  f ’(x) < 0
Sketching the derivative
Sketch the derivative of the function f(x) whose graph is shown.
At SP’s f ’(x) = 0
y = f ’(x)
Inceasing  f ’(x) > 0 y x
(1,5) 4 2 -1
(3,-3)
Deceasing  f ’(x) < 0
Now draw smooth curve through shaded areas and zero on x-axis