teoremavalormedioderivadasparte1

7/25/2019 TeoremaValorMedioDerivadasParte1

1/8

1

a G (x)

G (x) =

x sin

1x

; x= 0

a ; x= 0

G (x) =

sin

1x

; x= 0

a ; x= 0

x

x179 + 1631 + x2 + sin2 (x)

= 119

f(x) = xn, nN

lmh0

(x + h)n xn

h =nxn1

an bn = (a b) an1 + an2b + + abn2 + bn1

f(x) = x2 ; x0

x ; x

0

f (0)

g (x) =

x2 sin

1x

; x= 0

0 ; x= 0

0. g (0) = 0

f(x) =

1cos(x)

x2

x2 + x + 12

x >0

x


2/8

lmx0

f(x)

f x >0

f(x) =

5 + a sen(3x)sen

(5x)

bx + 5

x2

x


3/8

n f(x) = 0 x n n + 1.

f(x) = x3

x + 3

f(x) = x5 + x + 1

f(x) = x2, f(x + h) f(x)

h

x= 5, h= 3

x= 5, h= 0, 1

lmh0 f(x + h) f(x)h

f(x) = x2

f(x) = ax2 + bx + c

f(x) = 1x

f(x) = x3

f(x) = sin(x)

f(x) = cos(x)

f(x) = 3

x + 3

f(x) = x3. f (1) f 32 f(x) = 3

x

f (x)

f (1)

f (2) (1, 3)

2, 32

x f (x)

f(x) = ax2 + bx + c, a, bR f(x) = x5

f(x) = 3

x

f(x) =

|x

|,

f (2)

f (0)


4/8

f : R R T >0 x0

fx0+T

2 =f(x0)

f : ]a, b[R x1, x2, . . . , xn]a, b[ x0]a, b[

f(x0) = f(x1) + f(x2) + + f(xn)

n

f : R{1, 1, 4} R R

f(x) = x2 + 1

x 1 +x3 + 1

x 4 + (x 1) (x 2) (x 3) (x 4)

x2 1x + 1

f

f

1, 1 4 f

x0 f(x) =

a2 x2 |x0|< a a, b c R

f(x) =

4x x0

ax2 + bx + c 0< x


5/8

n=2. n=1

13

10, 1

2x

2ax + b

1x2

3x2

cos(x)

sin(x)

13

3x+3x+3

f (1) = 3, f 32 = 6, 75 3

x2

f (1) =3, f (2) = 34 y=3x + 6

y=34x + 3

2ax + b

5x4

1

3 3x2

f(x) =|x| ,

1

0

g: R R

g (x) = f

x +

T

2

f(x)

g

g (0) =fT

2 f(0)

g

T

2

=f(T) f

T

2

=f(0) f

T

2

f(0) = f

T

2

g (0) gT

2

x0

0, T2

g (x0) = 0

m = mn {f(x1) , f(x2) , . . . , f (xn)}M = max {f(x1) , f(x2) , . . . , f (xn)}

m M f(x1) , f(x2) , . . . , f (xn)

m f(x1) + f(x2) + + f(xn)n

M

x0


6/8

f

x= 1

lmx1+

f(x)

= lmx1

x2 + 1x 1 + x

3 + 1x 4 + (x 1) (x 2) (x 3) (x 4)

x2 1x + 1

= +

x= 4

lmx4+

f(x)

= lmx4+

x2 + 1

x 1 +x3 + 1

x 4 + (x 1) (x 2) (x 3) (x 4)

x2 1x + 1

= +

x=1

lmx1 f(x)

= lmx1

x2 + 1

x 1 +x3 + 1

x 4 + (x 1) (x 2) (x 3) (x 4)

x2 1x + 1

= 241

f(1) =241

lmxx0

f(x) f(x0)x x0 = lmxx0

a2 x2

a2 x20

x x0= lm

xx0a2 x2 a

2 x20x x0

1

a2 x2 +a2 x20=

x0a2 x20

x0 < 0

lmh0

f(x0+ h) f(x0)h

= lmh0

4 (x0+ h) 4x0h

= 4

0< x0 < 1

lmh

0

a (x0+ h)2

+ b (x0+ h) + c

ax20+ bx0+ c

h= lm

h0a (x0+ h)

2 + b (x0+ h) ax20 bx0h

= lmh0

(h (b + 2ax0+ ah))

h= b + 2ax0

x0 > 1

lmh0

f(x0+ h) f(x0)h

= lmh0

(3 2 (h + x0)) (3 2x0)h

= 2


7/8

x0 = 0 x0 = 1 x0 = 0

lmh0

f(0 + h) f(0)h

= lmh0

f(h)

h

lmh0+

f(h)

h = lm

h0f(h)

h

lmh0+

ah2 + bh + c

h = lm

h04h

h

lmh0+

ah2 + bh + c

h = 4

f 0 0

lmx0+

ax2 + bx + c

= lm

x04x= 0

c= 0

4 = lmh0+

ah2 + bh + c

h = lm

h0+ah2 + bh

h =b

lmh0

f(1 + h) f(1)h

= lmh0

f(1 + h) 1h

lmh0+

f(1 + h) 1h

= lmh0+

3 2 (1 + h) 1h

=2

lmh0

f(1 + h) 1h

= lmh0

a (1 + h)2

+ b (1 + h) + c 1h

= lmh0

a (1 + h)2

+ 4 (1 + h) 1h

x= 1

lmx1

ax2 + bx + c

= lm

x1+(3 2x) = 1

a + b + c= 1

a + 4 = 1

a=3

lmh0

3 (1 + h)2 + 4 (1 + h) 1h

=2

f(x) =

4x x03x2 + 4x 0< x


8/8

xf(a) af(x)x a =

xf(a) af(a) + af(a) af(x)x a

= (x a) f(a) a (f(x) f(a))

x a= f(a) a

f(x) f(a)

x a

lmxa

xf(a) af(x)x a = lmxa

f(a) a

f(x) f(a)

x a

=f(a) af (a)

f(x) g (a) f(a) g (a) + f(a) g (a) f(a) g (x)x a

= g (a)f(x) f(a)x a f(a)

g (x) g (a)x a

lmxa

f(x) g (a) f(a) g (x)x a

= lmxa

g (a)

f(x) f(a)

x a

f(a)

g (x) g (a)x a

= g (a) f (a) f(a) g (a)

teoremavalormedioderivadasparte1

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