Maximum and minimum of function on f (x) = Xe ^ - x The maximum and minimum of F (x) = Xe ^ - x in the interval [0,2] How to find the derivative of fxe ^ - x? Isn't e ^ - x itself a composite function?

Maximum and minimum of function on f (x) = Xe ^ - x The maximum and minimum of F (x) = Xe ^ - x in the interval [0,2] How to find the derivative of fxe ^ - x? Isn't e ^ - x itself a composite function?

f(x)'=e^(-x)-xe^(-x)
=e^(-x)(1-x)
So when x is on [0,1], f increases and on [1,2], f decreases
And f (0) = 0, f (1) = e ^ (- 1), f (2) = 2E ^ (- 2)
Therefore, the maximum value is e ^ (- 1), and the minimum value is 0
Supplement: derivation
f(x)'=[xe^-x]'
=(x)'e^-x+x(e^-x)'
=e^-x+xe^-x*(-x)'
=e^-x-xe^-x

The sum of the maximum and minimum values of the function y = 2 ^ x + log2 (x + 1) in the interval [0,1]

Function 2 ^ x is an increasing function in [0,1], log2 (x + 1) is also an increasing function, so the maximum value is y = 2 + 1 = 3 when x = 1, and the minimum value is y = 1 + 0 = 1 when x = 0

The minimum value of function f (x) = Xe ^ - x in interval [0,4] is

F '(x) = e ^ (- x) - Xe ^ (- x) = (1-x) · e ^ (- x) Let f' (x) = 0, the solution is x = 1. ① when 0 ≤ x < 1, f '(x) > 0, f (x) is an increasing function, and the minimum value is f (0) = 0. ② when 1 < x ≤ 4, f' (x) < 0, f (x) is a decreasing function, and the minimum value is f (4) = 4E ^ (- 4). Because f (0) < f (4), the minimum value is f (0) = 0

Find the maximum and minimum values of the function FX = x / X-1 in the interval [3,4]

The maximum value of 1 / F (x) = (x-1) / x = 1-1 / X in the interval [3,4] is 1-1 / 4 = 3 / 4, and the minimum value is 1-1 / 3 = 2 / 3,
So the maximum value of F (x) in the interval [3,4] is 3 / 2, and the minimum value is 4 / 3

The process of finding the maximum and minimum value of the function FX = X-1 / X in the interval (2.5)!

What are the maximum and minimum values of the function y = 5-2x-x ^ in the interval [- 2.1]?

y=5-2x-x^2=6-(x+1)^2
The function is a parabola with an opening downward, monotonically increasing on (- 1, - 1) and monotonically decreasing on (- 1, ∞)
When x = - 1, the function has a maximum value of y = 6
In the interval [- 2,1], the function takes the minimum value y = 2 at x = 1

The increasing interval of function y = - cos (1 / 2x - π / 3) is (); when x ∈ (0, π), the range of function is ()

y=-cos（1/2x-π/3)
y ' = - sin(1/2x - π/3)*(1/2)>0
We get: sin (1 / 2x - π / 3)

If | x | ≤ π / 4, then the maximum value of the function f (x) = cos ^ 2x + SiNx is

F(x)=cos^2x+sinx=-(sinx)^2+sinx+1
|X | ≤ π / 4 gives - √ 2 / 2

The range of y = (SiNx) 2 + sinx-1

Y = (SiNx) 2 + sinx-1 = (1 / 2 + SiNx) ^ 2-5 / 4 ∵ - 1 ≤ SiNx ≤ 1 ∵ - 1 / 2 ≤ 1 / 2 + SiNx ≤ 3 / 2 ∵ 0 ≤ (1 / 2 + SiNx) ^ 2 ≤ 9 / 4 ∵ - 5 / 4 ≤ (1 / 2 + SiNx) ^ 2-5 / 4 ≤ 1 ∵ y = (SiNx) 2 + sinx-1 range: [- 5 / 4,1]

Finding the range of y = Tan + 1 / tan-1

Let t = tanx-1, so y = 1 + 2 / T. the range of R2 / T is (- ∝, 0) ∪ (0, + ∝), so the range of Y is (- ∝, 1) ∪ (1, + ∝)