Extrablad användning av f(x)

(1)

Stockholms Tekniska Gymnasium 2013-03-19

Extrablad användning av f(x) Beräkna f(0)då

x x f( )=

1 )

(x = x+ f

1 )

(x = x− f

x x f( )=3

27 ( =x) f

x x

f( =) 2 x x

f( =) 5 x x

f( =) 11 x x f( )=

) 5

(x x

f =

11 2

7 )

(x x x

f = − +

x x

f 1

) ( =

Beräkna f(1)då x

x f( )=

5 )

(x = x+ f

1 )

(x = x− f

x x f( =) 2

x x

f( = ) 3 x x

f( =) 2,5 x x

f( )= 4 11 2

7 )

(x x x

f = − +

x x

f 1

) ( =

Beräkna f(2)då x

x f( )=

2 )

(x = x+ f

2 )

(x = x− f

x x f( =) 9

42 ) (x = f

x x

f( =) 2 x x

f( =) 9 x x

f( )= 18 2 )

(x = x² + f

x x

f 1

) ( =

1 )

(x = x+

f beräkna

) 0 ( f

) 1 ( f

) 2 ( f

) 6 ( f

) (−1 f

) (−2 f

) (a f

x x

f( )=3 beräkna )

0 ( f

) 1 ( f

) 2 ( f

) 5 ( f

) (−1 f

) (−3 f

) (a f

x x

f( )= beräkna )

0 ( f

) 1 ( f

) 4 ( f

) 36 ( f

) 25 , 0 ( f

) 36 , 0 ( f

1 2 )

(x = x−

f beräkna

) 0 ( f

) 1 ( f

) 2 ( f

) 5 ( f

) (−1 f

) (−2 f

) (a f

) 2

(x x

f = beräkna )

1 ( f

) 2 ( f

) 5 ( f

) (−1 f

) (−2 f

) (−5 f

2 )

(x = x² +

f beräkna

) 1 ( f

) 2 ( f

) 5 ( f

) (−1 f

) (−2 f

1 2 )

(x = x−

f beräkna

) (a f

) (a² f

) ( +a 1 f

) 1 (a² + f

) 1 ( h f +

) (a h

f +

) 1 ( ) 1

( h f

f + −

) 3 ( ) 3

( h f

f + −

) 2

(x x

f = beräkna )

(a f

) (a² f

) ( +a 1 f

) 1 (a² + f

) 1 ( h f +

) (a h

f +

) 1 ( ) 1

( h f

f + −

) 3 ( ) 3

( h f

f + −

1 2 )

(x = x+

f och

) 2

(x x

g = beräkna )

1 ( ) 1

( g

f +

) 2 ( ) 2

( f

g +

) 3 ( ) 5

( g

f +

) 1 ( ) 1

( g

f −

) 1 ( ) 1

( f

g −

) 2 ( ) 3

( g

f −

) 1 ( ) 1 ( g

f ⋅

) 3 ( ) 0

( g

f ⋅

) 3 ( ) 0

( f

g ⋅

)) 1 ( (g f

)) 1 ( ( f g

)) 0 ( (g f

)) 0 ( ( f g

(2)

Facit Beräkna f(0)då

0 )

(x = x= f

1 1 0 1 )

(x = x+ = + = f

1 1 0 1 )

(x = x− = − =− f

0 0 3 3 )

(x = x= ⋅ = f

27 ( =x) f

1 2 2 )

(x = ^x = ⁰ = f

1 5 5 )

(x = ^x = ⁰ = f

1 11 11 )

(x = ^x = ⁰ = f

0 0 )

(x = x = = f

0 0 )

(x = x⁵ = ⁵ = f

11 0

11 0 7 11

7 )

(x = x− +x² = ⋅ − + ² =− f

x x

f 1

)

( = Ej definierat. Division med 0 är inte tillåtet

Beräkna f(1)då 1 )

(x = x= f

6 5 1 5 )

(x = x+ = + = f

0 1 1 1 )

(x = x− = − = f

2 1 2 2 )

(x = x= ⋅ = f

3 3 3 )

(x = ^x = ¹ = f

5 , 2 5 , 2 5 , 2 )

(x = ^x = ¹ = f

2 4 1 4 4

)

(x = x = ⋅ = = f

3 11 8 1 11 7 1 11 1 7 11

7 )

(x = x− +x² = ⋅ − + ² = − + = − =− f

1 1 1 ) 1

( = = =

x x f

Beräkna f(2)då 2 )

(x = x= f

4 2 2 2 )

(x = x+ = + = f

0 2 2 2 )

(x = x− = − = f

18 2 9 9 )

(x = x= ⋅ = f

42 ) (x = f

4 2 2 )

(x = ^x = ² = f

81 9 9 )

(x = ^x = ² = f

6 36 2

18 18

)

(x = x = ⋅ = =

f

6 2 4 2 2 2 )

(x = x² + = ² + = + = f

5 , 2 0 1 ) 1

( = = =

x x f

(3)

1 )

(x = x+

f beräkna

1 1 0 ) 0

( = + =

f

2 1 1 ) 1

( = + = f

3 1 2 ) 2

( = + =

f

7 1 6 ) 6

( = + =

f

0 1 1 ) 1

(− =− + = f

1 1 2 ) 2

(− =− + =− f

1 )

(a = a+ f

x x

f( )=3 beräkna 0 0 3 ) 0

( = ⋅ = f

3 1 3 ) 1

( = ⋅ = f

6 2 3 ) 2

( = ⋅ = f

15 5 3 ) 5

( = ⋅ = f

3 ) 1 ( 3 ) 1

(− = ⋅ − =− f

9 ) 3 ( 3 ) 3

(− = ⋅ − =− f

a a a

f( )=3⋅ =3

x x

f( )= beräkna 0 0 ) 0

( = =

f

1 1 ) 1

( = =

f 2

4 ) 4

( = =

f

6 36 ) 36

( = =

f

5 , 0 25 , 0 ) 25 , 0

( = =

f f(0,36)= 0,36 =0,6

1 2 )

(x = x−

f beräkna

1 1 0 2 ) 0

( = ⋅ − =− f

1 1 2 1 1 2 ) 1

( = ⋅ − = − = f

3 1 4 1 2 2 ) 2

( = ⋅ − = − = f

9 1 10 1 5 2 ) 5

( = ⋅ − = − = f

3 1 2 1 ) 1 ( 2 ) 1

(− = ⋅ − − =− − =− f

5 1 4 1 ) 2 ( 2 ) 2

(− = ⋅ − − =− − =− f

1 2 1 2 )

(a = ⋅a− = a− f

) 2

(x x

f = beräkna 1

1 ) 1

( = ² = f

4 2 ) 2

( = ² = f

25 5 ) 5

( = ² = f

1 ) 1 ( ) 1

(− = − ² = f

4 ) 2 ( ) 2

(− = − ² = f

25 ) 5 ( ) 5

(− = − ² = f

(4)

2 )

(x = x² +

f beräkna

3 2 1 2 1 ) 1

( = ² + = + = f

6 2 4 2 2 ) 2

( = ² + = + = f

27 2 25 2 5 ) 5

( = ² + = + = f

3 2 1 2 ) 1 ( ) 1

(− = − ² + = + =

f f(−2)=(−2)² +2=4+2=6

1 2 )

(x = x−

f beräkna

1 2 )

(a = a− f

1 2 )

(a² = a² − f

1 2 1 2 2 1 ) 1 ( 2 ) 1

(a+ = a+ − = a+ − = a+ f

1 2 1 2 2 1 ) 1 ( 2 ) 1

(a² + = a² + − = a² + − = a² + f

1 2 1 2 2 1 ) 1 ( 2 ) 1

( +h = +h − = + h− = h+ f

1 2 2 1 ) ( 2 )

(a+h = a+h − = a+ h− f

h h

h f

h

f(1+ )− (1)=2(1+ )−1−(2⋅1−1)=2+2 −1−2+1=2 h h

h f

h

f(3+ )− (3)=2(3+ )−1−(2⋅3−1)=6+2 −1−6+1=2

) 2

(x x

f = beräkna ) 2

(a a

f =

4 2 2

2) ( )

(a a a

f = =

1 2 )

1 ( ) 1

(a+ = a+ ² =a² + a+ f

1 2 1

2 ) ( ) 1 ( ) 1

(a² + = a² + ² = a² ² + ⋅a² + =a⁴ + a² + f

2

2 1 2

) 1 ( ) 1

( h h h h

f + = + = + +

2 2

) ( )

(a h a h a ah h

f + = + = + +

2 2

2

2 (1 ) 1 2 1 2

) 1 ( ) 1 ( ) 1

( h f h h h h h

f + − = + − = + + − = +

2 2

2

2 (3 ) 3 2 3 9 9 6 9 6

) 3 ( ) 3 ( ) 3

( h f h h h h h h h

f + − = + − = + ⋅ ⋅ + − = + + − = +

1 2 )

(x = x+

f och g(x)= x² beräkna 4 1 1 2 1 1 1 2 ) 1 ( ) 1

( + g = ⋅ + + ² = + + = f

9 4 1 4 2 1 2 2 ) 2 ( ) 2

( + f = ⋅ + + ² = + + = g

20 9 1 10 3 1 5 2 ) 3 ( ) 5

( + g = ⋅ + + ² = + + = f

2 1 1 2 ) 1 ( 1 1 2 ) 1 ( ) 1

( − g = ⋅ + − ² = + − = f

2 3 1 1 2 1 ) 1 1 2 ( 1 ) 1 ( ) 1

( − f = ² − ⋅ + = − − = − =− g

3 4 7 4 1 6 ) 2 ( 1 3 2 ) 2 ( ) 3

( − g = ⋅ + − ² = + − = − =

f

3 1 ) 1 2 ( ) 1 ( ) 1 1 2 ( ) 1 ( ) 1

( ⋅ g = ⋅ + ⋅ ² = + ⋅ = f

9 9 1 ) 3 ( ) 1 0 2 ( ) 3 ( ) 0

( ⋅ g = ⋅ + ⋅ ² = ⋅ = f

0 7 0 ) 1 3 2 ( ) 0 ( ) 3 ( ) 0

( ⋅ f = ² ⋅ ⋅ + = ⋅ = g

3 1 2 1 1 2 ) 1 ( ) 1 ( )) 1 (

(g = f ² = f = ⋅ + = + = f

9 3 ) 3 ( ) 1 1 2 ( )) 1 (

(f =g ⋅ + = g = ² = g

1 1 0 2 ) 0 ( ) 0 ( )) 0 (

(g = f ² = f = ⋅ + = f

1 1 ) 1 ( ) 1 0 2 ( )) 0 (

(f =g ⋅ + =g = ² = g