Let f x be continuously derivable of the second order on 0,1, and f 0 f 1 , and f x M, prove that f

Review:It isLagrangetype of remainderTaylor's formula。It is not limited, and here eta and xi represent the Lagrangian remainder terms after only the first order, which is the f(x) straight equal sign.

He now brings in x=0 after point t and changes t back to x.

Math:

Mathematics is the study of concepts such as quantity, structure, change, space, and information. Mathematics is a general means for humans to strictly describe the abstract structure and pattern of things, and can be applied to any problem in the real world, and all mathematical objects are inherently artificially defined. In this sense, mathematics belongs to the formal sciences.

And not natural science. Different mathematicians and philosophers have a range of opinions on the exact scope and definition of mathematics.

This is Taylor's formula for the remainder term of the Lagrangian type.

This is the expression of the first-order Taylor Lagrangian remainder form, which is a rule. As for why two different symbols are used, it is because the domains are different. Then the back takes the absolutely worthwhile position and uses the two scaling is |a+-b|<=|a|+|b|And the title gives f''(x)<=m。

I feel that the hardest thing to think of is this scaling, which is a good question.

Although it is confirmed that the demand curve arises naturally from the theory of consumer choice, the derivation of the demand curve itself is not a theory that proposes consumer behavior. Simply determining how people react to change doesn't require a rigorous analytical framework. However, consumer choice theory is extremely useful.

As we will illustrate in the next section, we can use this theory to more deeply determine the factors that determine family behavior.

Instant Answer Draw budget constraint lines and indifference curves for Pepsi and pizza. Illustrate what happens to the budget constraint line and consumer optimum when pizza** rises. Use your graph to divide this change into income effects and substitution effects.

Four applications. Now that we've established the basic theory of consumer choice, we can now use it to illustrate four questions about how the economy works. However, since each problem involves family decision-making, we can address these issues with the consumer behavior model we just proposed.

Are all demand curves sloping to the bottom right?

In general, when an item** goes up, people buy less. Chapter 4 refers to this normal behavior as the law of demand. This pattern manifests itself in the demand curve sloping to the lower right.

However, as far as economic theory is concerned, the demand curve also sometimes slopes to the upper right. In other words, consumers sometimes defy the law of demand and buy more when one item rises. To illustrate how this can happen, look at Figure 21-12.

In this example, the consumer buys two items – meat and potatoes. Initially, the consumer budget constraint line was a straight line from A to B. The best advantage is c.

When the potato ** rises, the budget constraint line moves inward and is now a straight line from A to D. Now the best thing is e. It is important to note that the rise in potatoes has led consumers to buy more potatoes.

Use the partial integration method.

(0,1)x(1-x)f"(x)dx (u= x(1-x) v'= f''(x) u' =1-2x v= f'(x)

x(1-x) f'(x) ]0,1) -0,1)(1-2x)f'(x)dx and set u1 = 1-2x v1 = f'(x) (u1)' 2 (v1)'=f(x)

0 - 1- 2x) f(x) (0,1) -2 ∫^0,1)f(x)dx

f(1) +f(0) -2 ∫^0,1)fx)dx

Where: [x(1-x) f'(x) ]0,1) denotes the function [x(1-x) f'(x)] in the value of x=1 minus its value in x=0.

It's similar elsewhere. ,1,Let f(x) have a second continuous derivative on [0,1], proving: 1,2)f(x)dx=1 2[f(1)+f(2)]-1 2 (1,2)(2-x)(x-1)f"(x)dx

Consider f(a)=minf(x)=-1 then f'The second-order Taylor formula at (a) = 0 and f(x) to x=a is f(x) = f(a) + f'(a)(x-a)+f''(r)/2!*(x-a) 2 r belongs to (a,x)x=0 and has f(0)=-1+f''(x1)/2!*A 2 i.e. f''(x1)=2 a 2x=1 has f(1)=-1+f''(x2)/2!

1-a) 2 i.e. f''.

Use the partial integration method.

0,1)x(1-x)f"(x)dx (u= x(1-x) v'= f''(x) u' =1-2x v= f'(x)

x(1-x) f'(x) ]0,1) -0,1)(1-2x)f'(x)dx and set u1 = 1-2x v1 = f'(x) (u1)' 2 (v1)'=f(x)

0 - 1- 2x) f(x) (0,1) -2 ∫^0,1)f(x)dx

f(1) +f(0) -2 ∫^0,1)fx)dx

Where: [x(1-x) f'(x) ]0,1) denotes the function [x(1-x) f'(x)] in the value of x=1 minus its value in x=0. It's similar elsewhere.

Are you satisfied with the above? ,3,It's so complicated I don't want to understand,1,Let f(x) have a second continuous derivative on [0,1], prove: (1,2)f(x)dx=1 key on2[f(1)+f(2)]-1 2 (1,2)(2-x)(x-1)f"(x)dx

Let f(x) have a second continuous derivative on [0,1], and prove that the state is bright: 1,2)f(x)dx=1 2[f(1)+f(2)]-1 2 (1,2)(2-x)(x-1)f"(x)dx

Use the partial integration method.

(0,1)x(1-x)f"(x)dx (u= x(1-x) v'= f''(x) u' =1-2x v= f'(x)

x(1-x) f'(x) ]0,1) -0,1)(1-2x)f'(x)dx and set u1 = 1-2x v1 = f'(x) (u1)' 2 (v1)'=f(x)

0 - 1- 2x) f(x) (0,1) -2 ∫^0,1)f(x)dx

f(1) +f(0) -2 ∫^0,1)fx)dx

Where: [x(1-x) f'(x) ]0,1) denotes the function [x(1-x) f'(x)] in the value of x=1 minus its value in x=0.

It's similar elsewhere. ,6,|f(1)-f(0)|+=f(1)-f(0)|+ The last inequality is because the maximum value of the quadratic function x 2+(1-x) 2 on [0 1] is 1,2, and let f(x) have a second continuous derivative on [0,1], proving: 0,1)f(x)dx=1 2 (f(0)+f(1))-1 2 0,1)x(1-x)f"(x)dx

(0,1) represents the integral on the (0,1) interval.

Summary. Consultation Records · On 2021-11-212), let the function f(x) be third-order derivable on between[0,1] and f(o)=f(1)=0, f(x)=x, f(x) proof.

I haven't learned that formula yet, I've solved this problem, thank you.

Okay, classmate.

Proof: Your question is incorrect, it should be: f(1)=1

This question examines the medium theorem and the Lagrangian median theorem!

f(x) is continuous on [0,1], and according to the mesovalue theorem, x1,x2 (0,1), such that:

f(x1)=1/3

f(x2)=2/3

f(x) is derivable in the interval (0,x1),(x1,x2),(x2,1), and is continuous in [0,x1],[x1,x2],[x2,1], according to the Lagrangian median theorem:

1∈(0,x1)

2∈(x1,x2)

3∈(x2,1)

Such that: f(x1)-f(0).

f'(ξ1)·(x1-0)

f(x2)-f(x1)=f'(ξ2)·(x2-x1)f(1)-f(x2)=f'( 3)· (1-x2) thus: 1 f'(ξ1)

x1-0)/f(x1)-f(0)

x1/(1/3)=3x1

1/f'(ξ2)

x2-x1)/f(x2)-f(x1)

x2-x1)/(1/3)=3x2-3x11/f'(ξ3)

1-x2)/f(1)-f(x2)

1-x2)/(1/3)=3-3x2

All of the above are added up:

1/f'(ξ1)

1/f'(ξ2)

1/f'(ξ3)

3x1+3x2-3x1+3-3x2=3

Done! I've thought about it for an afternoon, so let's add a few points!

To use the cubic Lagrangian median theorem, the graph made me write the name backwards.