Ask the math emperor to solve these three high math problems, and ask to do the high math problem ma

1, let g(x)=f(x)=2x, then g(x) c[0,1] and g(0)=f(0)+0=0 g(1)=f(1)+2=3

According to the mesovalue theorem there is y [0,1] such that g(y)=1 (because 1 [0,3]).

So there is y [0,1] such that f(y) = 1-2y

2, the title is misprinted, it should be f(a)+2f(b).

Consider f(b) f(a).

then [f(a)+2f(b)] 3 [f(a),f(b)].

Because f(x) c[0,1].

So according to the mesovalue theorem, there is y [a,b] such that f(y) = [f(a)+2f(b)] 3

So there is y [a,b] such that 3f(y) = f(a)+2f(b).

3, because f(x) c[0,2] and f(0)+2f(1)=6

So according to the previous question, there is y [0,1], such that 3f(y)=f(0)+2f(1)=6 i.e. f(y)=2

Because f(2)=2 and f(x) is derivable on (0,2).

So according to Rolle's theorem, there exists z(y,2) 0,2), such that f'(z)=0

1, let f(x)=f(x)-(1-2x), and then use the intermediate value theorem;

2. Doubt the topic, but let =a can; ,3, first of all, it is determined that there is a point a in [0,1] in [0,1] such that f(a)=2 is generated, and then Rohr's theorem is used.

Answer: (1), f(0) = 0+k = 0+1, so, k = 1, Answer: d

2) When x tends to 0, ln(x+1) tends to ln1 and tends to 0, so the answer: c

3) Let h=1 m, m tends to infinity, h tends to 0

The original limit becomes lim[f(a+h)-f(a)] h = f'(a), so, the answer: d

4) Answer: C

Because after sin is derived four times, it is still sin, sin(x 2) is multiplied by a 1 2 for each derivative.

Because it is a chain derivative of the composite function, it is multiplied by 20 1 2.

5), x in x. The first derivative is 0 and the second derivative is also 0, so, in x. f(x) = constant.

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Can you write it out Your font is not flattering.

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Give me 200 copies first, I'll do it for you, the first question is derivable = differentiable, that is, as long as the derivability is proved, that is, the inverse value = the value of the function.

The first let t=e x tdx=dt t the original integral becomes 1 (t(t+1))dt= dt t- dt (t+1).

It's just 2 ln functions.

The second one is sin2x 2 on top

Here is (1+(sinx) 2)(1-(sinx) 2)=-(-2-2(sinx) 2)(2-2(sinx) 2) 4

(cos2x-3)(cos2x+1)/4

The original formula is - 2sin2xdx [(cos2x-3)(cos2x+1)].

dcos2x/[(cos2x-3)(cos2x+1)]

The following practice is the same as the first question, which is also 2 ln functions.

The third order e 1 2=t dx=2dt t

The original formula is 2dt ((t 2)(t+1))=2[ dt t 2 - dt t + dt (t+1)].

Set of formulas to figure it out.

1.Write the function of x corresponding to y and the first derivative y', which is the slope of its tangent. Since the point (0,1) is not on the curve, the tangent point (x0,y0) needs to be set by the tangent equation y-y0=y'(x-x0) and the curve equation to solve the tangent point, the reverse slope and the tangent equation (the curve equation is so complicated, find it yourself).

2.It is the area enclosed by the three by applying the definite integral. s1 = definite integral (from 1 2 to 2) 4x; s2 = definite integral (1 2 to 2) 1 x, the area enclosed by the three is s1-s2

3.Find Extremum: Find the first derivative y of y for the function', another y'=0, find the corresponding x value, and check y'monotonicity, judging whether the point is a maximum or minimum.

Find the inflection point: first find y', and y'Coiled; Seek y''(2nd derivative), another y''=0 if y'''If it is not equal to 0, then it is another y''The point of =0 is the inflection point of the curve.

Brother, these are the most basic problems of linear algebra, can you use the emperor of mathematics?