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L’Hospital’srule.The other indeterminate form.



Answer:

Theorem. Suppose that f and g are differentiable and g’ has no zeros on (a;b)

(x)= (x)=0 (1) or

(x)= (x)= (2)

And suppose that =L (3) Then =L (4)

Proof:

Suppose,that ε>0. From (3), there is x0 (a;b) such that | – L |<ε if x0<c<b(5)

Cauch’s Theorem implies that if x and t are in [x0,b) , then for every c between them , and therefore in (x0,b) , such that

(g(x)-g(t))f’(c)=(f(x)-f(t)g’(c)) (6)

Since g’ has no zeros in (a,b) Lagranzh’s Theorem implies that g(x)-g(t) 0 if x,t (а,b)

This means that g cannot have more than one zero in (a,b). Therefore, we can choose x0 so that , in addition to (5), g has zero in [x0,b)

Then (6) can be rewritten as = ,

so implies that as | -L|<ε if x,t (a,b) (7)

If (1) holds, Let x be fixed in [x0,b) , and consider the function

10.2 G(t)= -L

From (1) (t)= (t)=0

So (t)= -L (8)

Since, І G(t) І<ε if t (x0,b) because of (7) ,(8) implies that І –L| ε.This holds for all x in (x0,b) , which implies (4)

L’Hospital’s rule используется : (0/0) and (&/&).The other indeterminate form:0&, &-&, 00,1&,&0. Ары каратай далелдеу керек еще!!!!»

a) The indeterminate form 0 : we say that a product fg is of the form 0 , as x->b,if of the factors approaches 0 and the order approaches as x->b- In this case, it may be useful to apply L’Hospital’s rule after writing f(x)g(x)= =( ), f(x)g(x)= (since one of the rations is the form ( ) and the other is of the form ( ))

Similar statements apply to limits as x->b+, x->b, x->

b)The indeterminate form : A difference (f-g) is of the form ( ) as x->b, if

(x)= (x)=

In this case reduced to a common dehominator.

c)The indeterminate form 00,1&, 0. In this three cases, we use formula UV=eVLnU (f(x))g(x)=eg(x)Lnf(x).

Taylor’s formula.

Answer:

A polynomial is of the form P(x)=Q0+Q1(x-x0)+…+an(x-x0)n (1)

where Q0,Q1,…Qn and x0 are constants.

In particular, a constant polynomial P(x)=Q0 is of degree zero, if Q0 0

If f is differentiable all x0 , then f(x)=f(x0)+f’(x0)(x-x0)+ (x-x0)

Where =0

The polynomial T1(x)=f(x0)+f’(x0)(x-x0) which is of degree 1 and satisfies T1(x)=f(x0) T1(x0)=f’(x0), approximates f so well near x0 that

=0 (2)

Now suppose that f has n derivatives at x0 and Tn is the polynomial of degree n such that

Tn(r)(x0)=f(r)(x0) 0 r n (3)

Tn is polynomial of degree n;

Tn(x)=a0+a1(x-x0)+…+an(x-x0)n (4)

Differentiating (4) yields Tn(r)(x0)=r!ar

So (3) determines ar uniquely as ar=f(r)(x0)/r!

Therefore,

Tn(x)=f(x0)+ (x-x0) + (x-x0)2 + …+ (x-x0)n = (x-x0)r

We callTn the n-th Taylor polynomial of f about x0 (xxx дегенимиз x3)

1. sinx=x – +– -– +….

2. cosx=1 – – + – – – + – - …

3. ln(1+x)=x – x2/2 + x3/3 – x4/4 + x5/5 - …

4. ex= 1+ x + x2/2! +x3/3! + x4/4! + …

5. (1+x)a=1 + ax + a(a-1)x2/2! + a(a-1)(a-2)x3/3! + … + xa

Analysis of function using the derivative plotting graph of function.

Answer:

To construct the graph of the function y=f(x) to find:

1. Domain and Range

2. Symmetries f is said to be an even function (if f(-x)=f(x)) , and is said to be an odd function (if f(-x)=-f(x))

3. Points of discontinuity. The line x=a is called a vertical asymptote of the curve y=f(x), if =

4. Asymptotes (oblique or slant)

Y=kx + b, k= , b= if k=0, then y=b is a horizontal asymptote.

5. X-intercepts (zeros of function and the region of constant sign)

6. Relative extrema (max and min).

If f’(x)>0, for every value ox x in (a;b) , then f is increasing on [a;b].

If f’(x)<0, for every value ox x in (a;b), then f is decreasing on [a;b].

If f has a relative extremum at x=x0 is a critical point of f; that is , either f’(x0)=0 or f is not differentiable at x0.

7. Concavity , inflection points.

a) If f’’(x)>0 , for every value of x, then f is concave up on that interval.

b) If f’’(x)<0 , for every x in the open interval , then f is concave down on that interval.

If a function f changes the direction of its concavity at the point

(x0 ,f(x0)) , then we say that f has an inflection point at x0

 

13. Integration. Indefinite integral.

Answer:

Definition. A function f is called an Antiderivative of a function f on the given open interval, if F’(x)=f(x) for all x in the interval.

For example, the function F(x)= x4 is an antiderivative of f(x)=x3 on the interval (- ;+ ), because for each z in this interval. F’(x)=( x4)’=x3=f(x).

However, F(x)= x4 is not the only antiderivativeof f. If we add any constant C to x4 , then the function x4 + C is also an antiderivativeof f on (- ;+ ).

Since, ( x4 + C) = x3 = f(x).

If F(x) is any antiderivative of f(x), then F(x) + C is also an antiderivative on that interval. The process of finding antiderivative is called integration.

= F(x) + C.

3 dx = x4/4 + C.

The expression is called an indefinite integral.

Equation (1) should be real as:

“The integral of f(x) with respect to x is equal to F(x) plus a constant. ”

Integration formulas:

1) =x + C; 9) =arcsin + C;

2) ndx= xn+1 /n+1 + C; 10) a2 +x2= arctg + C;

3) =sinx + C; 11) =ln|x + | + C;

4) =-cosx + C; 12) a2 -x2= ln| + C;

5) cos2x=tgx + C; 13) = ln|x| + C;

6) sin2x=-ctgx + C; 14) = + a2/2arcsin + C;

7) xdx = ex + C; 15) = + a2/2ln|x + | + C.

8) xdx = ax/lna + C;





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