Why is the function f (x) = ax ^ 3 + BX ^ 2 + CX (a ≠ 0) odd, B must be 0?

Why is the function f (x) = ax ^ 3 + BX ^ 2 + CX (a ≠ 0) odd, B must be 0?

Odd function then - f (x) = f (- x)
f(x)+f(-x)=0
So ax & sup3; + BX & sup2; + CX + a (- x) & sup3; + B (- x) & sup2; + C (- x) = 0
ax³+bx²+cx-ax³+bx²-cx=0
2bx²=0
This is an identity
That is to say, it holds no matter what value x takes
So B = 0

Function f (x) = x ^ 2 (e ^ X-1 + ax + b) given that x = - 2 and x = 1 are the zeros of y = f '(x), find the values of a and B

f'(x)=2x(e^(x-1)+ax+b)+x^2(e^(x-1)+a)
f'(-2)=-4(e^(-3)-2a+b)+4(e^(-3)+a)=0
f'(1)=2(e^0+a+b)+e^0+a=0
8a-4b+4a=0
b=3a
2+2a+2b+1+a=0
2b+3a+3=0
6a+3a+3=0
a=-1/3
b=3a=-1

Given the number of zeros of function f (x) = LNX + ax in the interval (1,2)

f‘(x)=1/x+a
If a > = 0, then f '(x) > 0, monotonically increasing, at most one zero point. Moreover, f (1) = a > = 0, f (2) = LN2 + 2A > 0, so there is no zero point in (1,2)
If a

What is the slope and extremum of function tangent?

In essence, these two concepts are local properties of functions. Furthermore, they are properties of functions near point x0. Furthermore, they are properties of functions in the neighborhood of point x0. Therefore, most of them are related to limits and derivatives. The derivative of a function at point XO is called the slope of the tangent of a function at point x0

LIM (x → 0) {[∫ (upper x lower 0) ln (cost) DT] / x ^ 3}

According to lobi's rule, = Lim ln (cosx) / (3x & # 178;) = Lim ln (1 + cosx - 1) / (3x & # 178;) = LIM (cosx - 1) / (3x & # 178;) [Equivalent Infinitesimal Substitution] = - LIM (1-cosx) / (3x & # 178;) = - LIM (X & # 178 / 2) / (3x & # 178;) [Equivalent Infinitesimal Substitution] = - 1 /

LIM (x → 0) ∫ ln (cost) DT / x ^ 3 = (where ∫ superscript x, subscript 0) had better have a process

We omit the following limit process x --- > 0 Lim ∫ [0 to x] ln (cost) DT / X & # 179; lobita's law = Lim ln (cosx) / (2x & # 178;) = Lim ln (cosx-1 + 1) / (2x & # 178;) Equivalent Infinitesimal Substitution: ln (1 + U) and u equivalent = LIM (cosx-1) / (2x & # 178;) Equivalent Infinitesimal Substitution: cosx-1 and - X & # 178

Given limx → 0, ∫ (upper limit x lower limit 0) (2x-t) ln (1 + T) DT / x ^ n = k, find n

Replace with lobida's law and equivalent infinitesimal!
The original formula = LIM (x → 0) [∫ (2x-t) ln (1 + T) DT] / x∧n
＝lim(x→0)［∫(2x－t)t］/x∧n
＝lim(x→0)x²/nx∧(n-1)
＝lim(x→0)x/nx∧(n－2)
＝lim(x→0)1/n(n-2)x∧(n－3)
＝k
So x Λ (n-3) must be a constant, so n = 3

Lim x-infinity (√ 3x ^ 2 + 1) / x + 1

If the root sign is 3, the lower order items can be reduced

Lim x tends to infinity (1-2 / x) ^ 3x
E ^ - 6 is calculated by the luobida rule

The original formula = e ^ (LIM (x - > ∞) 3x * (- 2 / x))
=e^-6

lim_ x->1 (x^3-3x^2+2)/(x^3-x^2-x+1)

lim_ x->1 (x^3-3x^2+2)/(x^3-x^2-x+1）
=
lim_ x->1 (x-1)(xx-2x-2)/((x-1)^2(x+1))
=
lim_ x->1 (xx-2x-2)/((x-1)(x+1))
=
infinite