Given the function f (x) = Xe power (E is the bottom of natural logarithm), find the extreme value of function f (x)

Given the function f (x) = Xe power (E is the bottom of natural logarithm), find the extreme value of function f (x)

f'(x)=e^x+xe^x=(1+x)e^x=0
x=-1
Therefore, when x = - 1, there is a minimum f (- 1) = - 1 / E

It is known that the function f (x) = the x power of E + TX (E is the base of natural logarithm) (1) when t = - E, find the monotone interval and extreme value of the function (2 if for any x ∈ (0,2)] The inequality f (x) > 0 is constant, and the value range of real number T is obtained

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Let the function FX be equal to x + C to the power of ex (E is the base of natural logarithm, and C belongs to R) to find the monotonic interval of F (x)

f'(x)=(e^x-e^x(x+c))/e^2x=(1-x-c)/e^x
When x

The derivative f '(x) ＜ 0 of function f (x) defined on (0, + ∞) is constant, and f (4) = 1. If f (x + y) ≤ 1, the minimum value of x2 + Y2 + 2x + 2Y is ___

∵ f '(x) ＜ 0 ∵ the function is a subtractive function on (0, + ∞)
∵f（x+y）≤1=f（4）
∴x+y≥4
Let C = x2 + Y2 + 2x + 2Y, then (x + 1) 2 + (y + 1) 2 = C + 2, which represents the square of the distance from the point on the feasible region to (- 1, - 1), and also represents a circle
When x + y-4 = 0 is tangent to such a circle, its radius is the smallest, that is, the distance from the point on the row field to (- 1, - 1) is the smallest
∴(|−1−1−4|
2)2=18=c+2∴c=16
So the answer is: 16

Given that the image of the differentiable function f (x) on R is shown in the figure, the solution set of inequality (x2-2x-3) f '(x) > 0 is () A. （-∞，-2）∪（1，+∞） B. （-∞，-2）∪（1，2） C. （-∞，-1）∪（-1，0）∪（2，+∞） D. （-∞，-1）∪（-1，1）∪（3，+∞）

It can be obtained from the image that when f '(x) > 0, the function f (x) is an increasing function, so the solution set of F' (x) > 0 is (- ∞, - 1), (1, + ∞),
When f '(x) < 0, the function f (x) is a subtractive function, so the solution set of F' (x) < 0 is (- 1,1)
Therefore, inequality f '(x) < 0 is equal to the solution set of inequality (x-1) (x + 1) < 0
According to the meaning of the question, inequality (x2-2x-3) f '(x) > 0 is equivalent to inequality (x-3) (x + 1) (x + 1) (x-1) > 0,
Therefore, the solution set of the original inequality is (- ∞, - 1) ∪ - 1, 1) ∪ (3, + ∞),
Therefore, D

Let f (x) be a differentiable function defined on R and satisfy f '(x) > F (x). For any positive number a, the following inequality is always true, Let f (x) be a differentiable function defined on R and satisfy f '(x) > F (x). For any positive number a, the following inequality is constant () A.f(a)e^af(0) C.f(a)f(0)/e^a

Constructor
g(x)=f(x)/e^x
Then G '(x) = [f' (x) * e ^ x-e ^ x * f (x)] / (e ^ x) ²
∵ f'(x)>f(x)
∴ g'(x)>0
‡ g (x) is an increasing function on R
∵ a>0
∴ g(a)>g(0)
That is, f (a) / e ^ a > F (0) / e ^ 0 = f (0)
∵ e^a>0
∴ f(a)>f(0)*e^a
Select B