Find the maximum and minimum values of the function f (x) = | x ^ 2-3x + 2 | on [- 10,10]

Find the maximum and minimum values of the function f (x) = | x ^ 2-3x + 2 | on [- 10,10]

The solution is f (x) = | x ^ 2-3x + 2 | = | (x-3 / 2) & # 178; - 1 / 4 | = | (X-2) (x-1)|
The maximum point of the known function can only be obtained at x = 3 / 2, x = - 10, x = 10, x = 1, x = 2
f（3/2)=1/4,f（-10）=132,f（10）=72,f（1）=f（2）=0
The maximum value is 132 and the minimum value is 0

A necessary and sufficient condition for the function f (x) = A / 3x ^ 3 + B / 2x ^ 2 + CX + D to be a decreasing function on R

ax^2+bx+c

The cubic function f (x) = A / 3x ^ 3 + B / 2x ^ 2 + CX + D (a) is known

If R is monotonically increasing, then f '(x) = ax ^ 2 + BX + C is always nonnegative,
So a > 0 and B ^ 2 - 4ac = (a + B + B ^ 2 / (4a)) / (B-A)
= (2a+b)^2/(4a(b-a)) =[(b-a)+3a]^2/(4a(b-a))
>=[ 4(b-a)* 3a]/(4a(b-a)) = 3.

Given the function f (x) = CX + 1 (0 ＜ C), 2 ^ (- X / 2) (C ≤ x ＜ 1) satisfies f (C & # 178;) = 9 / 81. Find the value of constant C
2. Solve the inequality f (x) > 2 / 8 + 1

1. Idea: 0

Let f (x) = CX + 1 (0 ＜ x ＜ C) or F (x) = 2 - (2 / C & # 178;) (C ≤ x ＜ 1) satisfy f (C & # 178;) = 9 / 8 1 to find C
2. Solve the equation f (x) > 2 / 8 + 1
C & # 178; the results in the two formulas must be different. Why do I bring in 1, C = 1 / 2
2 does not solve one of the equations in 2

Because x = C ^ 22 ^ (1 / 2) / 4 is the unique solution of the inequality

The function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC about X is known, and its derivative is f '(x). Let g (x) = lf' (x) L,
Given the function f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BC, its derivative is f '(x). Let g (x) = | f' (x) |, note that the maximum value of function g (x) in the interval [- 1,1] is m
(1) If the function f (x) has a limit value of - 4 / 3 at x = 1, try to determine the value of B and C,
(2) If | B | > 1, it is proved that for any C there is m > 2
(3) If M > = k holds for any B, C. try to find the maximum value of K
My first question is b = 1, C = - 1, right?

Let me have a try... After sleeping... (1) f (x) = - 1 / 3x ^ 3 + BX ^ 2 + CX + BCF '(x) = - X & # - 178; + 2bx + C by the question, f' (1) = - 1 + 2B + C = 0f (1) = - 1 / 3 + B + C + BC = - 4 / 3bC + B + C + 1 = (B + 1) (c + 1) = 0, B = - 1 or C = - 1 if B = - 1, - 1-2 + C = 0, C = 3 if C = - 1, - 1 + 2b-1 = 0, B = 1, so (B, c) = - 1,3 or (1

Let the solution set of the inequality loga {2-1 / 2 (x ^ 2)} > loga (A-X) about X be a, and the intersection of a is Z = {1}. Then, the value range of constant a is obtained?

When 0 ＜ a ＜ 1, the inequality can be reduced to 2-1 / 2 x & sup2; > 0 → - 2 ＜ x ＜ 2 （1）2-1/2 x²＜a- x → x²-2x+2a-4＞0……… (2) The solution of inequality (2) is Δ = 4 – 4 (2a-4) > 0, that is, the solution is 1 - √ (5-2a) ＜ x ＜ 1 + √ (5-2a) because 0 ＜

Given a > 1, the inequality 2loga (x-1) > loga [1 + a (X-2)] about X is solved
Here a is the base, and the one in brackets is the exponent

That is, the square of (x-1) is greater than (1 + a (X-2)
That is, x square - (a + 2) x + 2A = 0, the solution is x = a, x = 2. When a > 2, then x > A or X

If the function g (x) = x3-ax2 + 1 is a monotone decreasing function in the interval [1,2], then the value range of real number a is______ ．

Let g (x) = x3-ax2 + 1, then G ′ (x) = 3x2-2ax, because & nbsp; G (x) = x3-ax2 + 1 is a monotone decreasing function in the interval [1,2], so g ′ (x) = 3x2-2ax ≤ 0 is constant in X ∈ [1,2]. That is to say, 2aX ≥ 3x2, a ≥ 32x is constant in X ∈ [1,2]

If the function f (x) = - x ^ 3 + ax ^ 2 + (A-1) X-1 is a monotone decreasing function on R, then the value range of real number a

f(x)=-x^3+ax^2+(a-1)x-1
f'(x)=-3x^2+2ax+(a-1)