It is proved that the function f (x) = X3 is an increasing function in its domain

It is proved that the function f (x) = X3 is an increasing function in its domain

Let x1

Prove that the function f (x) = √ X-1 / X is an increasing function in the domain of definition?

If your 1 / X is not in the root,
Let X1 > X2, f (x1) - f (x2) = √ x1-1 / X1 - √ x2 + 1 / x2
=√x1-√x2+(x1-x2)/x1x2
=(√x1-√x2)(1+(√x1+√x2)/x1x2),
Front (√ X1 - √ x2)

To judge the boundedness of a function, does function boundedness mean that it has both upper and lower bounds or only upper bounds?

Bounded. Upper bound is bounded, lower bound is bounded, both upper bound and lower bound are bounded

In higher numbers, if a function is bounded, does it mean that both its upper and lower bounds are equal?

Your understanding is wrong
The necessary and sufficient condition for boundedness is that there are both upper and lower bounds
It should be clear that the upper bound and the lower bound are not unique (the more detailed concept is the supremum supF (x), infimum INFF (x))

F (x) = x / 1-x & sup2; the function a with definition field () has upper bound but no lower bound B has lower bound but no upper bound C has bound D has neither upper bound nor lower bound

If there is no upper bound or lower bound, choose D
Because when x tends to 1 from the left, f (x) tends to positive infinity
When x tends to 1 from the right, f (x) tends to negative infinity

Is there an upper bound for a bounded function? I mean, can there be only an upper bound without a lower bound or can there be a lower bound without an upper bound?

Let f (x) be a function on the interval E. if for any x belonging to e, there are constants m and m such that m ≤ f (x) ≤ m, then f (x) is said to be a bounded function on the interval E. where m is called the lower bound of F (x) on the interval E and M is called the upper bound of F (x) on the interval E
This is the definition, needless to say, obviously not

Let x 1x 2 be a quadratic equation of one variable with respect to X. x (square) - 2 (m-1) x + m + 1 = 0. Two real roots are obtained. Y = x 1 + x 2 (both have squares). The analytic expression and domain of definition of y = f (M) are obtained

Weida theorem X1 + x2 = 2 (m-1) x1x2 = m + 1y = X1 & sup2; + x2 & sup2; = (x1 + x2) & sup2; - 2x1x2 = 4m & sup2; - 8m + 4-2m-2 = 4m & sup2; - 10m + 2 has roots, then the discriminant is greater than or equal to 04 (m-1) & sup2; - 4 (M + 1) ≥ 0m & sup2; - 3M ≥ 0m (M-3) ≥ 0m ≤ 0, m ≥ 3, so y = f (m) = 4m & sup2; - 10m + 2

It is known that the square of the quadratic equation x with respect to X - (square of M + 3) x + 1 / 2 (square of M + 2) = 0. Let x1x2 be two parts of the equation

x^2-(m^2+3)x+1/2(m^2+2)=0
x1+x2=m^2+3
x1*x2=1/2（m^2+2）
x1^2+x2^2-x1x2=17/2
(x1+x2)^2-3x1x2=17/2
(m^2+3)^2-3*1/2(m^2+2)=17/2
2m^4+9m^2-5=0
(2m^2-1)(m^2+5)=0
m=√2,m=-√2

Given the function f (x) = asin (2x + b) (a > 0), and f (x + Pie / 12) = f (Pie / 12-x) for any real number x, then the value of F (Pie / 3) is?

F (x) satisfies f (x + π / 12) = f (π / 12-x) for any real number X. it shows that f (x) takes x = π / 12 as the symmetry axis, that is, f (x) reaches the highest or lowest point at x = π / 12. Therefore, when x = π / 12, 2x + B = k π + π / 2 (K ∈ z), the solution is b = k π + π / 3 (K ∈ z). Therefore, f (π / 3) = asin (2 π / 3 + K π + π / 3) = as

F (x) = (x ^ 4 / 12) - (MX ^ 3 / 6) - (3x ^ 2 / 2) Q: if the real number m satisfies | m | ≤ 2, the function f (x) is always a "convex function" on (a, b), find B-
f(x)=(x^4/12)-(mx^3/6)-(3x^2/2)
Q: if the real number m satisfies | m | ≤ 2, the function f (x) is always "convex" on (a, b). To find the maximum absolute value of B-A, my answer is 4, but the correct answer is 2
Why can't we think that the maximum absolute value of B-A must be a and B are the two focuses of convex function and X-axis? In other words, a and B are the two roots of x ^ 2-mx-3 = 0, a + B = m, ab = - 3, absolute value B-A = root sign (a + b) ^ 2-4ab, and then the solution is 4?

f(x)=(x^4/12)-(mx^3/6)-(3x^2/2)
f'(x)=1/3x³-m/2x²-3x
f''(x)=x²-mx-3
∵ when the real number m satisfies | m | ≤ 2, the function f (x) is always "convex" on (a, b)
For any | m | ≤ 2, if x ∈ (a, b), f '' (x)