When x goes to 0, find the square of x times the limit of one square of SiNx The picture is wrongly typed and tends to 0

When x goes to 0, find the square of x times the limit of one square of SiNx The picture is wrongly typed and tends to 0

When x approaches 0, sin (1 / x) = 1 / x, so similarly, the above formula = 1

The function f (x) = AX2 + (B-8) x-a-ab is known. When x ∈ (- 3,2), f (x) > 0. When x ∈ (- ∞, - 3) ∪ (2, + ∞), f (x)

It can be seen conditionally that the two of a < 0 and f (x) = 0 are - 3 and 2
Then - 3 + 2 = - (B-8) / a... (1)
(-3)*2=(-a-ab)/a……(2)
From the solutions of (1) and (2), a = - 3, B = 5
Because AX2 + BX + C ≤ 0 is constant on [1,4]
That is - 3x ^ 2 + 5x + C ≤ 0 is constant on [1,4]
Then C ≤ 3x ^ 2-5x is constant on [1,4]
Let g (x) = 3x ^ 2-5x
Then G (x) = 3 (X-5 / 6) ^ 2-25 / 12
G (x) decreases monotonically on (- ∞, 5 / 6] and increases monotonically on [5 / 6, + ∞)
Then the minimum value of G (x) on [1,4] is g (1) = - 2
Therefore, C ≤ - 2

Let the two zeros of the function f (x) = ax ^ 2 + (B-8) x-a-ab) be - 3 and 2 respectively (1) Find the analytical formula of function f (x) (2) When the definition field of function f (x) is [0,1], find the value field of F (x)

-3 and 2 are the two roots of the equation AX ^ 2 + (B-8) x-a-ab = 0
-(B-8) / a = - 3 + 2 = - 1, the solution is B-8 = a
(- a-Ab) / a = - 1-B = - 3 * 2 = - 6, the solution is b = 5; Substitute the lion above to know a = - 3
So f (x) = - 3x ^ 2-3x-12
two
F (x) = - 3 (x ^ 2 + X + 4) the axis of symmetry is x = - B / 2A = - 1 / 2, which is not in the interval [0,1], so the function is monotonic in [0,1]
f（0）=-12 f(1)=-18
Therefore, the value range of the function in [0,1] is [- 18, - 12]

It is known that the two zeros of the function f (x) = AX2 + (B-8) x-a-ab are - 3 and 2, respectively (I) find f (x); (II) when the definition field of function f (x) is [0,1], find the value field of function f (x)

(1) By the meaning of the title
-b-8
a=-3+2
-a(1+b)
a=-3 × two ，
Solution
a=-3
b=5 ，
∴f（x）=-3x2-3x+18．
（II）f（x）=-3(x+1
2)2+3
4 + 18, monotonically decreasing at [0, 1],
∴f（1）=12≤f（x）≤f（0）=18，
The value range of the function is [12, 18]

Let the two zeros of function f (x) = a x2 + (B-8) (x-a - AB) be - 3 and 2 respectively, and find the analytical test of function f (x)

Let f (x) = a x2 + (B-8) (x-a - AB) = a (x + 3) (X-2), that is, ax ^ 2 + (B-8) x - (B-8) (a + AB) = ax ^ 2 + ax-6a. Compare the coefficients on both sides to get B-8 = a - (B-8) (a + AB) = - 6a to find a and B. A = (7 + √ 105) / 2, B = (23 + √ 105) / 2 or a = (7 - √ 105) / 2, B = (23 - √ 105) / 2 to find the solution of function f (x)

Function f (x) = AX2 = B, f (1 = - 1), f (2) = 8. (1) find the value of AB, (2) f (5) F (x) = AX2 = B, that 2 is a small 2. The computer can't type it out

f(x)=ax^2+b f(1)=-1 f(2)=8
Bring (1. - 1) (2.8) into the solution of the equation
a+b=-1
4a+b=8
a=3,b=-4
F (x) = ax ^ 2 + B into a, B
F (x) = 3x ^ 2-4 when x = 5, f (5)
=71