Lecture 3 - Trial and Improvementand Calculus I (Differentiation)

C2 Foundation Mathematics (Standard Track)

Dr Linda Stringer Dr Simon Craikl.stringer@uea.ac.uk s.craik@uea.ac.uk

INTO City/UEA London

Lecture 3 skills

I evaluate a functionI show that an equation has a solution between two given

valuesI use trial and improvement to find the solution to an

equation to 1 d.p.

I differentiate a polynomial expressionI find the gradient of a curve at a given pointI find a point on a curve with a given gradient

Lecture 3 vocabulary

I functionI polynomialI curveI tangentI differentiateI gradient function

Functions

I A function is a maths machine that takes a variable andreturns an expression involving that variable.

I For examplef (x) = x2

takes the variable x and returns x2.f (0) = 02 = 0× 0 = 0f (5) = 52 = 5× 5 = 25f (0.7) = 0.72 = 0.7× 0.7 = 0.49

I We can put other variables or expressions into a function.f (y) = y2

f (2x) = (2x)2 = 2x × 2x = 4x2

f (y − 1) = (y − 1)2 = (y − 1)(y − 1) = y2 − 2y + 1

Calculator skills

Consider the following five functions

f (x) = x5 − 2

g(x) = (x − 2)5

h(x) =√

x

p(x) =√

x + 1

q(x) = 5√

x

Use your calculator to evaluate

f (3), g(0.5), h(12), p(1

14), q(32)

Give your answer to 2 d.p. if it is not an integer

Graphical solution of equationsThe solutions to the equation f (x) = 0 are the x-intercepts ofthe graph of y = f (x)

For example, let f (x) = x2 − 4x + 3.The solutions to x2 − 4x + 3 = 0 are the same values of x aswhere the graph y = x2 − 4x + 3 crosses the x-axis.They are x = 1 and x = 3

1 2 3 4−1

1

2

3

y = x2 − 4x + 3

• • x

y

Showing that there is a solution between a and bQuestion: Show that there is a solution to x5− 8x + 2 = 0 in therange 1 < x < 2

−2 −1 1 2

−10

−5

5

10 y = x5 − 8x + 2

••

(1,−5)

•(2,18)y

15 − 8× 1 + 2 = −5 < 0 and 25 − 8× 2 + 2 = 18 > 0The graph of y = x5 − 8x + 2 crosses the x-axis between 1 and2, so there is a solution to x5 − 8x + 2 = 0 for some value of xbetween 1 and 2.(You can see from the graph that there is also another solutionbetween −2 and −1, and another solution between 0 and 1)

Showing that there is a solution between a and b

I To show there is a solution to f (x) = 0 for some value of xbetween a and b, first substitute a and b into f (x)

I If f (a) < 0 < f (b) or f (b) < 0 < f (a), then there is asolution to f (x) = 0 for some value of x between a and b

I Question: Show that there is a solution to x5 − 8x + 2 = 0for some value of x between 1 and 2

I Answer: Let f (x) = x5 − 8x + 2Then f (1) = 15 − (8× 1) + 2 = −5and f (2) = 25 − (8× 2) + 2 = 18f (1) < 0 and f (2) > 0 so there is a solution to f (x) = 0 forsome value of x between 1 and 2

Showing that there is a solution between a and b

Example: Show there is a solution to the equationx4 − 10x3 + 16x2 − 4x = 0 between 1 and 2

Trial and improvement

I Trial and improvement is a numerical method which we useto find a solution (to 1 d.p.) to an equation, when we knowthat a solution exists between two integers, a and a + 1

I Choose the midpoint of a and a+1 as your first value to try.I Repeatedly narrow down where the solution lies by trying

other (hopefully better) values between a and a + 1I Repeat until you know that the solution is between two

values that differ by 0.1I Substitute in the midpoint of these values to find out to

which value the solution is closest.

Trial and improvement - example 1Question: There is a solution to the equation x3 − 2x − 25 = 0between 3 and 4. Use trial and improvement to find this solutionto 1 d.p.

3.1 3.2 3.5

10

20

30

•

•

••

y = x3 − 2x − 25

x

y

Trial and improvement - example 1

I Question: There is a solution to the equationx3 − 2x − 25 = 0 between 3 and 4. Use trial andimprovement to find this solution to 1 d.p.

I Let f (x) = x3 − 2x − 25

x f(x)>0 or f(x)<0? Solution between3 33 − (2× 3)− 25 = −4 < 04 43 − (2× 4)− 25 = 31 > 0 3 and 43.5 3.53 − (2× 3.5)− 25 = 10.88 > 0 3 and 3.53.2 . . . =1.37 >0 3 and 3.23.1 . . . =-1.41<0 3.1 and 3.23.15 . . . = -0.04 <0 3.15 and 2

The solution is 3.2 to 1 d.p.

Trial and improvement - example 2

Question: The equation x5− 8x + 2 = 0 has a solution between1 and 2. Find this solution to 1 d.p.

21.5 1.7

−5

5

10

•

•

•

•

•

y = x5 − 8x + 2

x

y

Trial and improvement - example 2

I Question: The equation x5 − 8x + 2 = 0 has a solutionbetween 1 and 2. Find this solution to 1 d.p.

I Answer: Let f (x) = x5 − 8x + 2.

x f(x)>0 or f(x)<0? Solution between1 . . . =-5<02 . . . =18>0 1 and 21.5 . . . = -2.41<0 1.5 and 21.7 . . . =2.60>0 1.5 and 1.71.6 . . . = -0.31<0 1.6 and 1.71.65 . . . = 1.03>0 1.6 and 1.65

The solution is 1.6 to 1 d.p.

Why use numerical methods ?

I We use numerical methods to solve equations when wecan not use algebra, or as an alternative to using algebra.

I For example to solve the equation

x5 − 8x + 2 = 0

I The numerical method in this module is TRIAL ANDIMPROVEMENT.

I Other numerical methods for solving equations include trialand error, and iteration (we do not study these inFoundation Mathematics).

Gradients of lines and curves

I The gradient of a straight line is the same everywhere onthe line.

x

y

I What is the gradient of a curve?

x

y

TangentI The tangent to a curve at a point is a line that touches the

curve at that point, but does not touch the curve anywhereelse near to that point.

I The gradient of the curve at a point is defined to be thegradient of the tangent at that point.

I The gradient of a curve varies as we move along the curve.

•

x

y

Differentiation

I Differentiation is a method for working out the gradient of acurve at a given point.

I For a function y = f (x) we use the notation

dydx

to denote the derivative of y with respect to x.I dy

dx is the same as y ′, f ′(x), and ddx (f (x))

Differentiation rulesI Let x and y be variables, and a,b,n,m be constantsI If y = xn, then

dydx

= nxn−1

I If y = axn, thendydx

= anxn−1

I If y = xn + xm, then

dydx

= nxn−1 + mxm−1

I If y = axn + bxm, then

dydx

= anxn−1 + bmxm−1

Differentiation - tricky examplesI If y = xn, then

dydx

= nxn−1

I If y = x, thendydx

= 1x1−1 = 1x0 = 1

I If y = 5x, thendydx

= 5× 1 = 5

I If y = 1, thendydx

= 0x0−1 = 0

I If y = 3, thendydx

= 3× 0 = 0

Differentiating polynomials - examples

I Question: Differentiate y = x4 + x3 + x2

I Answer: dydx = 4x3 + 3x2 + 2x

I Question: Differentiate y = x2 + x + 1I Answer: dy

dx = 2x + 1I Question: Differentiate y = 5x3 + 7x2

I Answer: dydx = 5× 3x2 + 7× 2x = 15x2 + 14x

I Question: Differentiate y = 4x2 + 3x + 9I Answer: dy

dx = 4× 2x + 3× 1 + 9× 0 = 8x + 3

Gradient of a curve

I If a curve has equation y = f (x), then y is a function of xand the derivative dy

dx is also a function of xI The derivative is also called the gradient functionI To work out the gradient of a curve y = f (x) at the point

when x = a, we evaluate dydx at x = a.

I We use the notation dydx (a).

The graph of y = x2

−2 −1 1 2

1

2

3

4

••

•

•

x

y

y = x2

Gradient of a curve

I Question: A curve has equation y = x2.What is the derivative (gradient function)?What is the gradient at the point (1,1)?What is the gradient at (0,0)?What is the gradient at (−1,1)?What is the gradient at (2,4)?What is the gradient at (3,9)?

I Answer: dydx = 2x.

To find the gradient at (1,1), substitute in x = 1 into thegradient function.dydx (1) = 2× 1 = 2.The curve with equation y = x2 has gradient 2 at (1,1).To find the gradient at (0,0), substitute in x = 0 into thegradient function.dydx (0) = 2× 0 = 0.The curve with equation y = x2 has gradient 0 at (0,0).

Examples

I What is the gradient of the curve y = 8x5 at x = 2?I Answer: We have dy

dx = 40x4.

I So dydx (2) = 40× 24 = 640.

I What is the gradient of the curve y = 3x3 − x2 at x = −1?I Answer: We have dy

dx = 9x2 − 2x.

I So dydx (−1) = 9× (−1)2 − 2× (−1) = 11.

The curve y = 13x3 + 2x2 − 12x

Question: For what values of x does this curve have gradient 0?

−10 −5 5

−50

50

100

x

y

Points with a given gradient

I Question: For what values of x does the curve withequation

y =13

x3 + 2x2 − 12x

have gradient 0?I Answer: Find the derivative.

dydx

= x2 + 4x − 12

I Solve x2 + 4x − 12 = 0.I This factorises as (x + 6)(x − 2) = 0.I The curve has gradient 0 at x = −6 and x = 2.