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

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.

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

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

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)

Exercise 5A, page 67 Questions 3 to 7 inclusive

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

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 = 32 + 3×3 + 5 Point of contact is (3 , 23) y = 23

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)

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

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

Increasing and Decreasing Functions dy dx f (x)

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

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

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

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) = – 6 + 24x – 24x2 Squared term can never be negative Consider the graph of y = – 6 + 24x – 24x2 f ’(x) = 0 when 2x – 1 = 0 ie x = 1/2 – 6 + 24x – 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

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) = – 6 + 24x – 24x2 f ’(x) = 0 when x = ½ Consider the graph of y = – 6 + 24x – 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

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

Stationary points dy dx f (x)

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

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

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

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

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 – 18 + 5 = – 4 ie 2x - 6 = 0 x = 3 So a stationary point occurs at (3, -4).

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   

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 × 3 + 10 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

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

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

     + + – 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

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.

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.

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

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

     + – + 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)

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

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 = 2 circles + curved area

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

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

Show that the area of the red triangle is given by: C F x D 6 E Find the greatest and least areas of the triangle BEF x B 8 A Area = Rectangle – 3 triangles

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 + 8 = 16 The largest area since all other x values  24 – something When x = 0, H(x) = 24

The following questions are taken from past papers

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.

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

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

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)

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