Let the function f (x) be continuous on the interval [a, b], differentiable in (a, b), and ∫ (a, b) f (x) DX = f (b) (B-A). It is proved that there is at least Let the function f (x) be continuous on the interval [a, b], differentiable in (a, b), and ∫ (a, b) f (x) DX = f (b) (B-A). It is proved that there is at least one point in (a, b) ζ, Make f '（ ζ)= 0 Reference. It seems that the teacher said to use the integral mean value theorem first and then Rolle's theorem.

Let the function f (x) be continuous on the interval [a, b], differentiable in (a, b), and ∫ (a, b) f (x) DX = f (b) (B-A). It is proved that there is at least Let the function f (x) be continuous on the interval [a, b], differentiable in (a, b), and ∫ (a, b) f (x) DX = f (b) (B-A). It is proved that there is at least one point in (a, b) ζ, Make f '（ ζ)= 0 Reference. It seems that the teacher said to use the integral mean value theorem first and then Rolle's theorem.

∫ (a, b) f (x) DX = f (b) - f (b), so ∫ (a, b) f (x) DX = f (b) (B-A) [f (b) - f (a)] / (B-A) = f (b) by lachlange's theorem, exists ξ Make: [f (b) - f (a)] / (B-A) = f（ ξ)ξ ∈(a,b)b> ξ> a=>f( ξ)= F (b) by L Rolle theorem, existence ζ ∈( ξ, b) Make f ′（ ζ)= 0 ζ ∈( ξ, b)=> ζ ∈...

Let the function f (x), when x is less than or equal to 0, be the (1-x) power of 2; When it is greater than 0, it is f (x-1). F (x) = x + A has and only two unequal real roots, then the value range of real a?

When x > 0, the image of F (x) = f (x-1) is equivalent to the function image 2 ^ (1-x) on the [- 1,0] interval   (- 1 ≤ x ≤ 1), see the figure below;
According to the image shape of the functions y = f (x) and y = x + A, when the straight line is in the area between the two red lines in the figure, f (x) = x + A has and only has two real roots; Beyond this range, the equation has either only one real root or at least three real roots;
The intercept a of the upper limit red line y = x + A on the Y axis is 4, and the corresponding equation is y < x + 4;
The lower limit red line passes through the point (1,4), Y-axis intercept a = 3, and the limit condition is Y > x + 3;
Therefore, x + 3 < x + a < x + 4, that is, 3 < a < 4;

Given the function f (x) = e ^ x-x ^ 2-1, it is proved that the equation f (x) = 0 has no negative real root by the method of counter proof

prove:
Suppose that the equation f (x) = 0 has negative real roots
Then x

Given that the function f (x) = the x power of A-2 + 1 / 1, it is proved that f (x) is always an increasing function no matter a is any real number

f(x)=1/(a-2^x+1)
f'(x)=2^x*ln2/(a-2x+1)^2
If 2 ^ x > 0 and LN2 > 0, it can be seen from the definition field that (a-2x + 1) ^ 2 > 0
Therefore, f '(x) > 0 means that f (x) increases monotonically

Given the function f (x) = the x-th power of a, G (x) = x - 2 / x + 1, it is proved that the equation f (x) + G (x) = 0 has no negative root (a greater than 1)

The counter proof method assumes that there is a negative number with - t > 0
Bring into equation a ^ (- t) + - t) - 2 / (- t) + 1 = 0
Sorted a = 1 / ((T-2 / t-1) ^ (1 / T))
∵a＞1 ,∴0＜((t-2/t-1)^(1/t)＜1
This inequality t has no solution, so there is no negative number
Take your time. Don't read it wrong. It's bad to read it wrong,

Find the derivative of the function: ① sin1 / x power of y = E. ② y = lncos (x power of E)

1.y=e^(sin 1/x) → y'=e^(sin 1/x)*cos(1/x)*(-x^(-2))
2.y=ln(cos(e^x)) → y'=1/(cos(e^x))*(-sin(e^x))*e^x=-sin(e^x)*e^x/(cos(e^x))