1. Derivative the following functions:

1. i) f (x) = x⁵ – 3 + 7

x2

ii)

f (x) = 2x -1

x + 1

iii)

f (x) = e^x ln x²

1. Solve the equations:

x² + 2x +1

1. 2x – 5

= x +1

1. ii) 2000 = 200e

0.025 x

1. Solve the diffrences:

1. i) 2x² – 35 > -9x

ii)

x² + 2x +1 >     +

2x – 5

1. Calculate the integrals:

1. i) 1 (x² + 3)dx

0

ii)

ò x²e^x3 dx

Oppgave 2

We have the function

f (x) = x³ + x² – x + 2

1. Show that f(-2) = 0 and use this to find other eventual zero-points for f(x)

Calculate f“(x) and show when f(x) grows and declines

1. Find eventual maximum and minimumspoints for
2. Make a sketch of the graph to f (x)

Oppgave 3

f (x) .

Remember to show the calculation, it’s not sufficient to show results on the calculator

The value of a apartment is 4 000 000. As a result of a longlasting

Verdien av en leilighet er 4 000 000. Som en følge av en langvarig recession the value sinks the apartment prices with 10% yearly, assume that the whole price-drop happens on the end of each year.

1. Whats the value of the apartment after 5 years?

1. How long does it take before the value of the apartment is halved?

You’re granted a loan for 3 000 000 to buy a apartment. The loan is a annuity loan (morgage) for 30 years with fixed yearly repayments, and the first repayment happens a year after the loan is granted.

The interest rate is 2% each year.

Renten er 2% per år.

1. How much is the yearly term amount (innstalment)?

1. Right after the 5th repayment, you decide to redeem the loan. What is the outstanding debt at this point?

Oppgave 4    (Vekttall 4)

The aviation company «Up in the Air» has estiamted that the demand for pyjamas at a chosen route (with A380) as a function of the price (P) is given by x(p) = 600 – p, where 0<p<600.

Flyselskapet Up in the Air har beregnet at etterspørselen etter pyjamaser på en bestemt flyrute

(med A380) som funksjon av prisen (p) er gitt ved

x( p) = 600 – p , der 0 < p < 600

1. Find the phrase/expression for price elasticity Elp x( p)

1. Calculate the price elasticity if the price is given by p=200, and give a interpretation of the result.

p = 200 , og gi en tolkning av

1. Assume that «Up in the air» wishes to increase their sale income by selling pyjamas. base it of the answer you got in task b) and explain why the company should increase, reduce or hold the price unchanged fra the point p=200

.

1. Calculate which price the demand is price-neutral, meaning that the price elascity is equal -1 . Explain why this price maximizes the sale-income.

Oppgave 5    (Vekttall 5)

A company sells the products A and B, where product A sells in x units and product B sells in y units. By sale of x units of A the compant achieves the price p per united given by

p = 400 -1.5x – y ,

And by sale of y units of product B, the company achieves the price q per unit given by

q = 450 – x – 2 y .

The costs of production by producing and selling x and y units of the two products is given by

C(x, y) = 100x + 50 y + 0.5x² + y² +10000 .

1. Show that the companys profit function is given with

f (x, y) = -2x² – 2xy – 3y² + 300x + 400 y -10000 .

.

1. Calculate the partially derivated first order for f(x,y) and find the only stationary point

1. Show that the stationary point to f(x,y) is a local maximumpoint, show the profit in this case.

1. Beacuse of capacity restrictions the compant must produce approximately 70 units totals of the two products. Calculate how many units of product A and B must be produced if the company wishes maximum profit

1. Whats the profit gicen in the answer of the previous question?

Oppgave 6 (Vekttall 4)

Under is shown the graph for the derivated function of f(x), i.e the graph of f ` (x). The function is defined for all realistic numbers, Df = R

1. Use the graph above to find the stationary points to f(x) and classify these. The answer shall be explained. It is not nessesary to show precisly where the stationary points are located – a approximate idication is sufficient

1. Use the graph above to find the turning point, or the turning points for the function f(x). Your answer must be explained. It is not nessesary to show precisly where the turning point or points are located – a approximate indication is sufficient

Formler til eksamen i MET09104 Matematikk for økonomer

Potenser:

aⁿ                          m                                                      1          ¹

an × an = an+m                         = an–m

am

(an )

= an×m

a– n =

an

a 2 =

(a × b)ⁿ = aⁿ × bⁿ

æ a ön

ç    ÷

è    ø

= an                                 =

bn

a ×                  =

Kvadratsetningene:

(a + b)² = a² + 2ab + b²

Annengradslikning:

(a – b)² = a² – 2ab + b²

(a + b)(a – b) = a² – b²

2                                                                                                                                                                                              2

ax + bx + c = 0 Þ x =

2a

som gir x1 og x2:

ax + bx + c = a(x – x1 )(x – x2 )

Lineære funksjoner:

y = ax + b           y – y

= a(x – x )

y – y

= y2 – y1 (x – x )

1                              1                                        1          x – x            1

2           1

Logaritmer og eksponensialuttrykk:

eln x = x

ln(a × b) = ln a + ln b

ln æ a ö = ln a – ln b

è    ø

ln a^x = x ln a

ln e = 1      ln1 = 0

y = e^x Û x = ln y

Aritmetiske rekker:

sn = a1 + a2 + ××× + an = a1 + (a1 + d ) + (a1 + 2d ) + ××× + (a1 + (n -1) × d )

s   = a1  + an × n       d = a – a         a = a + (i -1) × d

n                  2                       i            i-1                i             1

Geometriske rekker:

s = a + a + ××× + a

= a + a × k + a × k ² + ××× + a × k^n–1

n             1            2

kⁿ -1

n             1           1                   1                                     1

a                       i-1

s  = a                 k =    ⁱ         a = a × k

^{n      1} k -1

i            1

i-1

s = a1 + a2 + a3 + ××××××××××××××××××××××××

s = a1

1- k

Annuiteter (Finansmatematikk): Sluttverdi rett etter siste innbetaling:

(1+ r)ⁿ -1

An   = K         r

(1+ r)ⁿ -1

Sluttverdi ett år etter siste innbetaling:

An   = K (1+ r)         r

Nåverdi når første tilbakebetaling skjer om ett år:

K0 = K

(1+ r)ⁿ -1 (1+ r)ⁿ × r

(1+ r)ⁿ × r

Fast årlig tilbakebetaling når første tilbakebetaling skjer om ett år:

K = K0 (1+ r)ⁿ -1

Derivasjon:

f (x) = xⁿ

f (x) = g(x)ⁿ

Þ f ‘(x) = n × x^n–1

Þ f ‘(x) = n × g(x)^n–1 × g ‘(x)

f (x) =                           Þ

f ‘(x) =    1

2 x

f (x) = k

f (x) = g(x) × h(x)

f (x) = g(x)

h(x)

f (x) = ln x

Þ f ‘(x) = 0

Þ f ‘(x) = g ‘(x) × h(x) + g(x) × h ‘(x)

Þ f ‘(x) = g ‘(x) × h(x) – g(x) × h ‘(x)

h(x)²

Þ f ‘(x) = 1

x

f (x) = ln g(x)

Þ f ‘(x) =

1

g(x)

g ‘(x)

f (x) = e^x

f (x) = e^g(x)

f (x) = a^x

Þ f ‘(x) = e^x

Þ f ‘(x) = e^g(x) × g ‘(x)

Þ f ‘(x) = a^x × ln a

Integrasjon:

xⁿ dx = 1 xⁿ⁺¹+C

n +1

n ¹ -1

1 dx = ln | x | +C x

   1

x + a

dx = ln | x + a | +C

ò e^x dx = e^x + C

e^ax dx = 1 e^ax + C a

ò ln x dx = x ln x – x + C

Klassifisering av stasjonære punkter for funksjoner av to variable:

A = fxx ”(x0 , y0 )

B =  fxy ”(x0 , y0 )      C =

f yy ”(x0 , y0 )

AC – B² > 0 og AC – B² > 0 og AC – B² < 0

A < 0  Þ

A > 0  Þ

Þ