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 U^V=e^VLnU (f(x))^g(x)=e^g(x)Lnf(x).

Taylor’s formula.

Answer:

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

where Q₀,Q₁,…Q_n 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 T_n is the polynomial of degree n such that

T_n^(r)(x₀)=f^(r)(x₀) 0 r n (3)

T_n is polynomial of degree n;

T_n(x)=a₀+a₁(x-x₀)+…+a_n(x-x₀)ⁿ (4)

Differentiating (4) yields T_n^(r)(x₀)=r!a_r

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

Therefore,

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

We callT_n 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. e^x= 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! + … + x^a

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 = e^x + C; 15) = + a²/2ln|x + | + C.

8) ^xdx = a^x/lna + C;