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Anonymous users2024-02-071.Solution: f(x-a)=x(x-a)=(x-a+a)(x-a).
So f(x)=x(x+a).
2.For the second question you wrote, I don't think ABC is correct, so the answer should be the D you didn't write.
Odd functions. Definition: Let the function y=f(x) be defined in the field d, if there is x d for any x in d, and f(-x)=-f(x), then this function is called an odd function.
1. In the odd function f(x), the signs of f(x) and f(-x) are opposite and the absolute values are equal, that is, f(-x)=-f(x), and conversely, the function y=f(x) that satisfies f(-x)=-f(x) must be an odd function. For example: f(x)=x (2n-1),n z; (f(x) is equal to x's 2n-1 power, n is an integer) 2. The odd function image is symmetrical with respect to the center of the origin (0,0).
3. The domain of the odd function must be symmetrical with respect to the center of the origin (0,0), otherwise it cannot be an odd function.
4. If f(x) is an odd function and x belongs to r, then f(0)=0
5. Let f(x) be derivable on i, and if f(x) is an odd function on i, then f'(x) is an even function on i.
That is, f(x)=-f(-x) is the derivative f for it'(x)=[-f(-x)]'(-x)'=-f'(-x)(-1)=f'(-x)
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Anonymous users2024-02-061. Let f(x-a)=x(x-a)(a>0 be constant), then f(x)= 2The odd function of the following functions is a, 1n(1+x) b, e negative x square c,x+cosx
Solution: (1). f(x-a)=x(x-a)=(x-a)²+ax-a²=(x-a)²+a(x-a)，∴f(x)=x²+ax
2).None of them are odd functions.
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Anonymous users2024-02-051、f[(x+a)-a]=(x+a)[(x+a)-a]=(x+a)x
2. There are no odd functions in the three answers you gave. Satisfied.
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Anonymous users2024-02-04ln(1+x), what kind of function is this?
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Anonymous users2024-02-03<> two digging and selling with Luo bend scattered reeds to bury the belt.
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Anonymous users2024-02-02<> is like Qiao, and Kaibu is noisy and filial piety.
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Anonymous users2024-02-01Find the differential equation y'xlnx-y=1+ln x satisfies y(e)=1.
Solution: Find homogeneous equation y first'General solution of xlnx-y=0:
Isolate the variables to get dy y=dx (xlnx).
The result of the integral lny= dx (xlnx) = d(lnx) (lnx)=lnlnx+lnc=ln(clnx).
Therefore y=clnx; Change c to the function u of x, and get y=ulnx....1)
Deriving (1) from both sides of x yields :y'=u/x+(lnx)u'...2)
Substituting (1) and (2) into the original formula yields: [u x + (lnx)u']xlnx-ulnx=1+ln²x
Simplify to get x(ln x)u'=1+ln²x
Separate the variables to obtain du=[(1+ln x) (xln x)]dx
The result of the integral is u= dx (xln x) + dx x= d(lnx) (ln x)+lnx=-1 lnx+lnx+c
Substituting equation (1) gives the general solution: y=lnx(-1 lnx+lnx+c)=-2+ln x+clnx
Then substitute the initial condition y(e)=1 to get 1=-2+1+c, so c=2
Therefore, the special solution that satisfies the initial condition is: y=ln x+2lnx-2
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Anonymous users2024-01-31an*2^n=(an*3^n)*(2/3)^n;This can be compared with a geometric progression, with a common ratio of 2 3;
Note that (an*3 n) is bounded and is derived from convergence.
Question 1: You can directly use Lobida's rule to directly derive the numerator and denominator of the previous test, and get f(x)=xf(x) (2*x), and then remove x, you can get f(x)=f(x) 2, and because f(0)=1, that is, f(x)=1 2;Since f(x) is continuous at x=0, i.e., a=1 2. >>>More
Since you said that it is the first semester of your junior year, then I advise you to focus more on professional courses, because professional courses also have to be studied well, and it is not too late to prepare for the next semester!!
I'd like to ask what the t in the first question is ...... >>>More
The first question is itself a definition of e, and the proof of the limit convergence can be referred to the pee. >>>More
An infinitesimal is a number that is infinitely close to zero, but not zero, for example, n->+, (1, 10) n=zero)1 This is an infinitesimal and you say it is not equal to zero, right, but infinitely close to zero, take any of the values cannot be closer to 0 than it (this is also the definition of the limit in the academic world, than all numbers ( ) are closer to a certain value, then the limit is considered to be this value) The limit of the function is when the function approaches a certain value (such as x0) (at x0). 'Nearby'The value of the function also approaches the so-called existence of an e in the definition of a value, which is taken as x0'Nearby'This geographical location understands the definition of the limit, and it is no problem to understand the infinitesimite, in fact, it is infinitely close to 0, and the infinitesimal plus a number, for example, a is equivalent to a number that is infinitely close to a, but not a, how to understand it, you see, when the chestnut n->+, a+(1, 10) n=a+ is infinitely close to a, so the infinitesimal addition, subtraction, and subtraction are completely fine, and the final problem of learning ideas, higher mathematics, is actually calculus, and the first chapter talks about the limit In fact, it is to pave the way for the back, and the back is the main content, if you don't understand the limit, there is no way to understand the back content, including the unary function, the differential, the integral, the multivariate function, the differential, the integral, the differential, the equation, the series, etc., these seven things, learn the calculus, and get started.