How much is 3 / 2 plus 2 / 3

How much is 3 / 2 plus 2 / 3

It's thirteen sixths,

If x and y satisfy (x-1) 2 + (y + 2) 2 = 4, find the maximum and minimum of S = 2x + y

(x-1) 2 + (y + 2) 2 = 4 denotes a circle with radius equal to 2 centered on (1, - 2). From S = 2x + y, y = - 2x + s is obtained. When the line is tangent to the circle, s obtains the maximum and minimum values. From | 2 × 1 − 2 − s | 22 + 12 = 2, s = ± 25, ｜ Smax = 25, smin = − 25

In the coordinate plane, if | x | + | y | = 1, then the maximum and minimum values of S = y-2x are

By drawing,
|x|+|y|=1
=> x+y=1(x>=0,y>=0)
x-y=1(x>=0,y

The odd function f (x) defined on [- 1,1] is known. When x ∈ [- 1,0], f (x) = 1 / 4x-a / 2x (a ∈ R)
A - f (- 1) = f (1) gives a = 1.7
If B F (0) = 0, then a = 1
X is the index
Which is right and why?

f(x)=1/4^x-a/2^x (x∈[-1,0]
f(x)=-f(-x)
=-[1/4^(-x)-a/2^(-x)]
=-4^x+a*2^x (x∈[0,1]
f(1)=-f(-1)
-4^1+a*2^1=-[1/4^(-1)-a/2^(-1)]
-4+2a=-4+2a
a∈R
We can't get the specific value of a, so it's wrong
∵x∈[-1,1]
∴f(0)=0
1/4^0-a/2^0=0
a=1
B is right

The odd function f (x) defined in [- 1,1] satisfies if x ∈ (0,1], f (x) = 2x / 4x + 1
The odd function f (x) defined in (- 1,1) satisfies if x ∈ (0,1), f (x) = 2x / 4x + 1
Finding the analytic expression of F (x) on (- 1.1)
、

When x ∈ (- 1,0)
Then - x ∈ (0,1)
So f (- x) = - 2x / (- 4x + 1)
Because f (x) is an odd function
So f (x) = 2x / (1-4x)
Since the function is odd, f (0) = 0
The analytic expression of F (x) on (- 1.1) is
f(x)=2x/(4x+1) x∈(0,1)
f(x)=2x/(1-4x) x∈(-1,0]

Given that f (x) is an odd function defined on R, when x ≥ 0, f (x) = 2x ^ 2-4x, find the function analytic formula of F (x)

Odd function f (x) = - f (- x)
When x ＜ 0, - f (- x) = - [2 (- x) ＾ 2-4 (- x)] = - (2x ＾ 2 + 4x) = - 2x ＾ 2-4x]

It is known that f (x) is an odd function defined on R, and if x > 0, f (x) = x & # 178; + X-1,
f(x)=____________ When x ＜ 0, f (x)=_____________ .

Fill in: 0 & nbsp; & nbsp; - X & # 178; + X + 1 respectively
The odd function x = 0 crosses the origin when it is defined
When x = 0, f (x) = 0
When x < 0, - X & gt; 0, then
f(x)=-f(-x)=-[(-x)²+(-x)-1]=-x²+x+1
That is, when x < 0, f (x) = - X & # 178; + X + 1

If the minimum value of quadratic function f (x) is 1 and f (0) = f (2) = 3 (1), find the analytic expression of F (x) (2) if f (x) is not monotone in the interval [2a, a + 1], find the value range of A
If the minimum value of quadratic function f (x) is 1 and f (0) = f (2) = 3 (1), find the analytic expression of F (x) (2) if f (x) is not monotonic in the interval [2a, a + 1], find the value range of a Hurry, hurry!

1. The minimum value is 1, that is, the vertex ordinate is 1,
Let f (x) = a (x-m) ^ 2 + 1 from F (0) = f (2) = 3
So am ^ 2 + 1 = 3
a(2-m)^2+1=3
M = 1, a = 2, that is, f (x) = 2 (x-1) ^ 2 + 1,
2. The axis of symmetry x = 1 of F (x) is in the interval [2a, a + 1], that is, 2A < 1 < A + 10 < a < 1 / 2

If the quadratic function f (x) = x ˇ 2-6x + 8, X ∈ [2, a] and the minimum value of F (x) is f (a), then the value range of a is process reason

The opening of this function is upward, and the axis of symmetry is x = 3

The quadratic function f (x) = 4 (x square) - 4ax + a square - 2A + 2 has a minimum value of 3 in the interval [2,2], and the value of a is obtained by finding a

Obviously, a ≠ 0, because the coefficient of the quadratic term is 4 > 0, the opening of the function image (parabola) is upward. If 3 is the minimum value of the function, it is obviously not in line with the meaning of the problem. Therefore, consider two cases where the function is an increasing function or a decreasing function in the interval [0,2], that is, f (0) = 3 or F (2) = 3, a = 1 ± 2 under the root sign or a = 5 ± 10 under the root sign, respectively