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L’Hospital’srule.The other indeterminate form.↑ ⇐ ПредыдущаяСтр 3 из 3 Содержание книги
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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) T’1(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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