How to prove that the third power of function FX = x - x is an odd function

How to prove that the third power of function FX = x - x is an odd function

∵f(x)=x^3-x
And f (- x) = - x ^ 3 + X
=-(x^3-x)
=-f(x)
That is, f (- x) = - f (x)
Ψ f (x) = x ^ 3-x is an odd function

For function f (x), if there exists x0 ∈ r such that f (x0) = x0 holds, then x0 is called the Tiangong No.1 point of F (x). It is known that the two Tiangong No.1 points of function f (x) = AX2 + (B-7) x + 18 are - 3 and 2 respectively. (1) find the value of a, B and the expression of F (x); (2) when the domain of definition of function f (x) is [T, t + 1] (T > 0), the maximum value of F (x) is g (T), and the minimum value of F (x) is g (T) (t) And find the expression of H (T) = g (T) - G (T)

(1) In this paper, f (- 3) = 3, f (2) = 2; that is, 9A + 21-3b + 21-3b + 18 = -3, 4A + 2b-14 + 18 = 2, and the solution is a = -3, B = 5 {f (x) = -3x2-2x + 18 (2) (f (3) = f (- 3) = 3 (3) = 3-3, f (2) = 3 (3) (f (2) = 3, f (3) (9a + 21-21-3b + 21-3b + 3B + 18 = -3, 9A + 21-3b-3b + 18-3b-3-3b-3b-3b-3-3b-3b-3b-3b + 1 (t [t, t + 1] (T > 0 [T, t [T, t, t + 1] (t [t [T, t + 1] (t (T) = f (T) = f (t (T) = f (t (T) max) max = f (x) max = f (g (T) - G (T) = 6T + 5

If f (x) = AX3 + bx2 + CX + D (a > 0) is an increasing function on R, then the relation of a, B, C is______ ．

If f ′ (x) = 3ax2 + 2bx + C ＞ 0, 4b2-12ac ＜ 0 is obtained, and B2 ＜ 3aC is simplified. Therefore, the answer is: a ＞ 0 and B2 ≤ 3aC

Lim xsin [ln (1 + 3 / x)] (x tends to infinity) =?
Senior one
Lim xsin [ln (1 + 3 / x)] (x tends to infinity) =?

Ln (1 + 3 / x) = ln [(1 + 3 / x) ^ (x / 3)] ^ (3 / x) = 3 / X ln (1 + 3 / x) ^ (x / 3) Lim xsin [ln (1 + 3 / x)] (x tends to infinity) = Lim x sin [3 / X ln (1 + 3 / x)] ^ (x / 3)] (x tends to infinity) = Lim xsin (3 / x) (x tends to infinity) = Lim 3 sin (3 / x)

lim(x-0)(x+1)ln(1+x)/x^2+3x
Using Equivalent Infinitesimal Substitution,

When X -- > 0, ln (x + 1) x
lim(x-->0) (x + 1)ln(x + 1)/(x² + 3x)
= lim(x-->0) [(x + 1)x]/[(x + 3)x]
= lim(x-->0) (x + 1)/(x + 3)
= (0 + 1)/(0 + 3)
= 1/3

LIM (x - > infinity) [ln (1 + 1 / x)] / [arc Cotx] using the law of lobida

LIM (x - > infinite) [ln (1 + 1 / x)] / [arc Cotx]
At the same time, we obtain LIM (x - > infinity) [ln (1 + 1 / x)] / [arc Cotx]
=LIM (x - > infinite) [[- 1 / (x ^ 2 + x)] / [- 1 / (1 + x ^ 2)] = LIM (x - > infinite) (1 + x ^ 2) / (x ^ 2 + x) = 1

LIM (x tends to 0) [∫ (from 0 to x) (1 + 2t) ^ (1 / Sint) DT] / ln (1 + x)

2e^2

LIM (x → 0) ∫ (upper limit x, lower limit 0) sin (T ^ 2) DT / x ^ 3
Limit process, 1 / 3

The definition of lobita rule is as follows:
 
 
 
If you don't understand this question, you can ask,

Lim x → 0 ∫ sin T ^ 2 DT / x ^ 3 =? From X product to 0

It is not necessary to find ∫ sin T ^ 2 DT
When x → 0, the numerator and denominator tend to 0, we use the lobita rule and the variable upper limit integral to obtain the derivative formula
=sinx^2/3x^2
Let x ^ 2 = t
When t approaches 0, the limit sin T / T = 1
So SiNx ^ 2 / 3x ^ 2 = 1 / 3
So the limit of the original formula is 1 / 3

Find LIM (x tends to 0) 1 / (x ^ 3) ∫ (upper limit is x, lower limit is 0) sin (T ^ 2) DT

Using the law of lobida, it's a good idea
=lim sin(x^2) / (3x^2)
=LIM (x ^ 2) / (3x ^ 2) [Equivalent Infinitesimal Substitution]
= 1/3