Find the function f (x) = x2−2x+2+ Minimum value of x2 − 4x + 8

Find the function f (x) = x2−2x+2+ Minimum value of x2 − 4x + 8

f(x)＝
(x−1)2+(0−1)2+
(x − 2) 2 + (0 − 2) 2 can be regarded as the sum of the distances from point C (x, 0) to point a (1,1) and point B (2,2), and be the point a '(1, - 1) symmetrical about the X axis of point a (1,1)
∴f(x)min＝
12+32＝
ten

Function f (x) = (x) ²+ The minimum value of 4x + 5) / (x + 1), (x > - 1) is

Let t = x + 1 > 0
Then x = T-1
Substitute f (x) = (T ^ 2-2t + 1 + 4t-4 + 5) / T = (T ^ 2 + 2T + 2) / T = 2 + T + 2 / T
t> 0, t + 2 / T > = 2 √ (t * 2 / T) = 2 √ 2. When t = 2 / T, that is, when t = √ 2 (x = √ 2-1 at this time), take the equal sign
Therefore, f (x) > = 2 + 2 √ 2, when x = √ 2-1, take the minimum value

Function f (x) = x ²- 4X + 6, X belongs to [1,5], find the value range of the function. F (x) = X ²+ BX + 1, minimum 0, B =?

From the topic, a = 1, B = - 4, C = 6, axis of symmetry x = - 2A / b = 2, so the minimum value f (2), the maximum f (5), axis of symmetry = - B, is substituted into the equation f (- b) = 0 to get B

Function y = 4x ²+ What is the minimum value of 8x + 13 \ 6 (1 + x)?

y=(4x^2+8x+13)/6(x+1)
6y(x+1)=4x^2+8x+13
4x^2+(8-6y)x+13-6y=0
If the equation has a solution, the discriminant > = 0
Namely: (8-6y) ^ 2-4 * 4 (13-6y) > = 0
64-96y+36y^2-208+96y>=0
36y^2>=144
y^2>=4
y> = 2 or Y

x> 0, find the function y = (x) ²- Minimum value of 4x + 1) / X

x>0
y=x-4+1/x≥2√(x*1/x)-4=-2
So the minimum value is - 2

Find the function y = √ (x ²＋ 1﹚＋√﹙x ²－ 4X + 8) and find the value of X at this time

Y = √ [(x-0) ^ 2 + (0 + 1) ^ 2] + √ [(X-2) ^ 2 + (0-2) ^ 2] so y is the sum of the distances from a point P (x, 0) to a (0, - 1) and B (2,2) on the X axis. Obviously, when APB is in a straight line and P is between AB, there is a minimum value AB on both sides of the X axis, and the qualified minimum value is the length of AB = √ [(0-2) ^ 2 + (- 1-2) ^ 2] = √ 13K (AB) = (