Ask for answers! Calculus Calculus Problems Ask for help!!!

Solution: (1) The image of f (x) and g (x) is symmetrical with respect to x = 1.

g(x)=f(2-x)

and when x 2,3], g(x) = a (x 2) 2 (x 2) (a is constant).

When x 2,3] f(2-x)=a(x-2)-2(x-2), i.e. f(x)=2x when 2-x [-1,0] and -x [-1,0] when x [0,1].

f(-x)=-2x³+ax

The function f(x) is an even function.

f(x)=f(-x)

f(x)=-2x³+ax x∈[0,1]

Therefore, the analytic formula of the function f(x) is.

f(x)=2x³-ax x∈[-1,0]

f(x)=-2x³+ax x∈[0,1].

2) The function f(x) is an increment on [0,1].

f'(x)=-6x +a 0 is constant on x [0,1].

Solution a 6.

3) When 0 a 6 f'(x)=-6x +a=0 gives x= (a 6).

When x = (a 6), take the maximum value -a 6* (a 6)*2+a* (a 6)=4

The solution is a = 6 (rounded).

In the same way, at [-1,0], a = -6 (rounded) is calculated when -6 a 0.

On [0,1] is a subtractive function and on [-1,0] is an increasing function.

The maximum value is f(0)=0 at x=0

Therefore, it is not satisfied with the topic.

In summary, a (a6,6) can not be made so that the maximum value of f (x) is 4.

Solution: (1) When -1 x 0, 2-x [2,3], and y=f(x) at any point p(x,y).

About the symmetry point p of the straight line x=1'(2-x,y) are all on the y=g(x) image

f（x）=g（2-x）=a（2-x-2）-2（2-x-2）^3=2x^3-ax

f(x) is an even function.

At 0 x 1, f(x)=ax-2x 3,2)0 x 1 f '(x)=a-6x 2>=0 a>=6x 2>=6 so a>=6

From f(x) is an even function, so only consider x [0,1], from f (x) = 0 to get x = root number (a 6), from f ( root number (a 6) ) ) = 6, where x = 1, when a (-6, 6), the maximum value of f(x) cannot be 4

The title of the first graph is as follows: just take the derivatives of x and y on both sides of the equation.

z'(x)cosz=yz+xyz'(x).So z'(x)=yz/(cosz-xy).

z'(y)cosz=xz+xyz'(y).So z'(y)=xz/(cosz-xy).

1. This question is a 0 0 type infinitive;

2. The method of solving the problem is as follows:

a. Factorization;

b. Robida's Derivative Rule.

3. The specific answers are as follows:

It's a matter of the language of calculus. The core of the calculus language is to use static processes to describe dynamic processes, and the limit can be understood as a "skill" of x, giving an "error", and then as long as x fully exerts this "ability", the "error" between f(x) and the limit can be sufficiently small. This is too abstract, so I'll talk about the first example.

When x approaches +, the limit of f(x) is a, which gives an "error" first, which is called , which is a requirement, and requires that the distance between f(x) and a be smaller than this; Because the limit trend of x is approaching +, the "ability" of x can be very large, how to show that the "error" can be small enough after x gives full play to the ability of "can be very large"? Let's find a standard n to measure the "ability", the characteristic of this n is "sufficiently large", if x is larger than n, the ability will be fully exerted, and then the error can be smaller. The specific statement is:

For any >0 (to give an error requirement at random), there is n>0 (a criterion for judging x ability can be found), so that when x > n (x gives full play to his ability according to this criterion) there is always |f(x)-a|< value between f(x) and the limit will satisfy the error requirement).

Similarly for any >0, there is n>0 such that when x<-n there is always a |f(x)-a|<ε

Note that it is not infinite, the ability of this x is not arbitrarily large, but arbitrarily close to x0, and the standard for measuring x's ability is a δ, as long as the distance between x and x0 is less than δ even if x gives full play to its ability, it can achieve a small distance ratio between f(x) and a. Tell whether to approach x0+ or x0 - it's a question of who minus whom)

For any >0, there is δ>0 such that when x-x0 <δ (x approaches x0 from the right, so x-x0) there is always a |f(x)-a|<ε

For any >0, there is δ>0 such that when x0-x <δ (x approaches x0 from the left, so x0-x) there is always a |f(x)-a|<ε

If you think about this yourself and understand this kind of question, you will have no problem at all. The above method of understanding is not what I thought, it was said by our physics teacher, and I just polished the language. I found his point very instructive, and he said that the language of calculus is actually a kind of idea similar to a physical experiment.

Physicists are concerned with the gap being small enough, not with the logical rigor of mathematicians; As a result, mathematicians also had a headache when they really needed to define the mathematical concept of "limit", and it took centuries to come up with the current language of calculus. We can also feel that analytics is not very profound, and its real ideological core is closely integrated with the real world. (The last paragraph is too philosophical, and if the landlord is not interested, I won't ......say it.)I like to ramble on my philosophical understanding when answering questions).

I really don't know about this! There should be example questions in the book, right?