Let f (x) have continuous derivatives on [0,1], and f (0) = f (1) = 0. It is proved that | ∫ (0,1) f (x) DX | ≤ 1 / 4max (0 ≤ x ≤ 1) | f '(x)|

Let f (x) have continuous derivatives on [0,1], and f (0) = f (1) = 0. It is proved that | ∫ (0,1) f (x) DX | ≤ 1 / 4max (0 ≤ x ≤ 1) | f '(x)|

The derivative f '(x) of F (x) is continuous on [a, b], and f (b) = a, f (a) = B. It is proved that the definite integral ∫ [a, b] f (x) f' (x) DX = 1 / 2 (a ^ 2-B ^ 2)

∫[a,b]f(x) f`(x)dx=
=(1/2)∫[a,b]df²(x)
=(1/2)f²(x)|(a,b)
=(1/2)(f²(b)-f²(a))
=(a²-b²)/2

It is proved that f (x) has continuous second derivative at x = 0
It is proved that the second derivative of F (x) is bounded

It is proved that the second derivative of F (x) is bounded

F (x) has second derivative on (0, + ∞). For all x ＞ 0, | f (x) | ≤ a, | f '' (x) | ≤ B, a, B are constants. It is proved that: | f '(x) | ≤ 2 √ ab

Let any positive number x and H have
f（x+h）=f(x)+f '(x)*h+1/2*f ''(x+θh)*(h^2)
Where 0

Let f (x) have continuous second derivative from negative infinity to positive infinity, and f (0) = 0, let g (x) = f (x) / x, X is not equal to 0; G (x) = a, x = 0
Determine the value of a so that G (x) is continuous from negative infinity to positive infinity

Answer: A & nbsp; = & nbsp; F & nbsp; & & nbsp; (0)
Obviously, when x ≠ 0, G (x) is continuous. Therefore, it is necessary and only necessary to be continuous when & nbsp; X & nbsp; = & nbsp; 0 & nbsp
According to the condition of continuity, the following formula must and only needs to be true

The sufficient condition for f (x) to reach the extreme value at x = C is that the first derivative is equal to 0 and the second derivative is not equal to 0. Why is this condition not a sufficient and necessary condition
Is it because some functions have no derivatives when they reach the extreme value?

Let me expand it for you. Why is this condition a sufficient condition? First of all, the sufficient condition is that the function is second-order differentiable. If for any n-order differentiable function, we can know by Taylor expansion, as long as the odd order derivative is equal to zero (all equal to zero), the even order derivative is not equal to zero (at least the second order derivative can't wait

Let the curve y = f (x) be tangent to the x-axis at the origin, the function f (x) has continuous second derivative, and when x ≠ 0, the first derivative of F is not equal to 0. It is proved that the curvature radius of the curve at the origin is r = limx → 0|x ^ 2 / (2f (x))|

Isn't that the curvature of a parabola

f(x)={sinx+1 x≥0,2x-1 x

That is to find the left and right limit of F (x),
lim(x→0+)f(x)
= sin0 +1
= 1
and
lim(x→0-)f(x)
= 2*0 -1
= -1

If f '(0) = 2, f (0) = 0, LIM (x tends to 0) f (2x) / SiNx is equal to

lim【x→0】f(2x)/sinx
=Lim [x → 0] f (2x) / x [Equivalent Infinitesimal Substitution]
=lim【x→0】[f(2x)-f(0)]/x
=2lim【x→0】[f(2x)-f(0)]/(2x)
=2f '(0)
=2×2
=4
Answer: 4

LIM (x-sinx) / [x ^ 2ln (1 + x)] x tends to 0

Taylor expansion
sinx=x - x³/3! + x^5/5! + ……
ln(1+x)=x - x²/2 + x³/3 + ……
The original formula = LIM (- X & # 179 / / 3! + x ^ 5 / 5! + )/( x³ - x^4/2 + x^5/3 + …… ) = -1/3