Why does the function f (x) = (2x-x Square) e have a maximum and no minimum

Why does the function f (x) = (2x-x Square) e have a maximum and no minimum

f(x)=(2x-x ²) e^x
f'(x)=(2-2x)e^x+(2x-x ²) e^x=(2-x ²) e^x
Let f '(x) > 0
So - √ 2

Given that the domain of function FX is (0,2], then the domain of function f √ x + 1?

The definition field of F (x) is (0,2],
that
In F (√ x + 1), the value range of √ x + 1 is (0,2],
therefore
0 solution - 1 < x ≤ 3
Therefore, the domain of this function is (- 1,3]

Please know that the definition field of function f (x) = root 3-x + (root x + 2) is set a, B = {x|x is less than a} It is known that the definition domain of function f (x) = root 3-x + (root x + 2) is set a, B = {x|x is less than a}

The definition field of function f (x) = 1 / 3 of root 3-x + (root x + 2) is set AA = {x | - 2 < x < = 3} B = {x | < a} if a is included in B, a is a subset of B, and the value range of a is a > 3. If a = - 1, find CRA = {x | - x < = - 2 or x > 3} CRB = {x | > = - 1} a ∩ (CRB) = {x | > 3}

Y = definition field of LG (3 under the root sign - (3 under the root sign and then minus 1) TaNx -- Tan square x), The title should say this: y = LG (root 3 minus (root 3 minus 1) TaNx minus the square of TaNx)

(root 3 minus (root 3 minus 1) TaNx minus the square of TaNx) is greater than 0, let Tan x = a
A square + (root sign 3-1) a-root sign 3 > 0, ten sub multiplication, (a + root sign 3) (A-1) > 0
So a > 1 or a < - radical 3, the solution is（ П/ 4+k П,П/ 2+k П) ∪（- П/ 2+k П,-П/ 6+k П), K is an integer

If the definition field and value field of function y = root X are [a, b], then a + B= A. 1 / 2 b. root 2 / 2 C.1 D.2

Y = √ x (x ≥ 0) is defined as [a, b]
Then the value field is [√ a, √ b]
∴a=√a,b=√b
The solution is a = 0, B = 1
‡ a + B = 1, select C

Known function y = x The domain of X − 1 is a, and the function y = 1 − x2 The value range of 1 + X2 is B (1) Find sets a and B; (2) Find a ∩ B, a ∪ B

(1) From the question: X ≥ 0 and X-1 ≠ 0, so a = {x | x ≥ 0 and X ≠ 1};
By y = 1 − x2
The solution of 1 + x2 gives x2 = 1 − y
Y + 1 ≥ 0, i.e. y − 1
Y + 1 ≤ 0, then Y-1 ≤ 0 and y + 1 > 0 or Y-1 ≥ 0 and y + 1 < 0, the solution is - 1 < y ≤ 1 or no solution
So B = {y | - 1 ＜ y ≤ 1}
(2) From a and B in (1), a ∩ B = [0, 1),
A∪B=（-1，+∞）．