High School Math Find the minimum value of 1 x 2 1 1 x 2 , x 0, . A2 2

y= (1+(x 2))+1+(1 (x 2))) let x=tana a (0, 2).

1+x 2=1+tan 2a=1 cos 2a1+1 x 2=1+cot 2a=1 sin 2aso. y=1/sina+1/cosa

sina+cosa)/sinacosa

The mean theorem sina + cosa> sinacosa=2 sinacosa sinacosa sinacosa

2/√sinacosa

2/[(2sinacosa)/2]

2 (sin2a 2) sin2a maximum value = 1, so the minimum value of the function is 2 2

(1+x^2)+√1+1/x^2)

Let x=tana (0=2 1 cosaxsina if and only if cosa=sina i.e. a= 4

So 1 cosa+1 sina>=2 1 1 2=2 2 if and only if x=1

There are 2 ways to solve this problem:

Method. 1. By the basic inequality: a+b 2 (ab).

Add 2 (ab) on both sides at the same time

Get: ( a+ b) 4 (ab).

So [ (1+(x 2))+1+(1 (x 2)))).

4 [(1+x)(1+1 x)]] if and only if 1+x = 1+1 x equal sign becomes immediate x=1).

4√(x²+1/x²+2)

4 (2+2) if and only if x = 1 x equal sign to immediate x = 1).

So (1+(x 2))+1+(1 (x 2))))2 2 (if and only if x=1 equal sign holds).

i.e. (1+(x 2))+1+(1 (x 2)))) minimum = 2 2

Method. 2. Let y= (1+(x 2))+1+(1 (x 2))) let x=tana a (0, 2) (think about why a (0, 2) is because x>0).

1+x²=1+tan²a=1/cos²a

1+1/x²=1+cot²a=1/sin²a

So. y=√(1+(x^2))+1+(1/(x^2)))

1/sina+1/cosa

sina+cosa)/sinacosa

From the fundamental inequality we get the sina+cosa 2 sinacosa

2 sinacosa sinacosa (if and only if sina = cosa equal sign to immediately a = 4, i.e. x = 1).

2/√(sinacosa)

2√2/√(sin2a)

2 2 (if and only if sin2a=1 equals to immediate a=4, i.e., x=1).

So y= (1+(x 2))+1+(1 (x 2))) minimum is 2 2 (if and only if x=1 equal sign holds).

Note: Fundamental inequalities are special Cauchy inequalities.

Because x 0, y 0 x+2y 5

So x 5-2y 0 y 5 2

i.e. 0 x 5, 0 y 5 2 , 0 xy 25 2(x+1)(2y+1) xy

2xy+x+2y+1）/√xy

2√xy+（x+2y+1)/√xy

2√xy+6/√xy

and xy minimum infinity approaches 0, so the minimum value of the original formula is 6

Solution: Let x+2y=t, x=t-2y

x²−xy+y²=1,(t-2y)²−t-2y)y+y²=1,7y²−5ty+t²-1=0

Discriminant =25t -4 7(t -1) 0-3t +28 0

t²≤28/3

2√21/3≤t-≤2√21/3

The value range of x+2y is [-2 21 3,2 21 3].

High School Mathematics: Let x 0, y 0, x+2y=5 find the minimum value of (x+1)(2y+1) xy?

Hello! Glad to answer for you!! Your sedan is coarse and pro.

Because x 0,y 0 x+2y 5 so x 5-2y 0 y envy 5 2 i.e. 0 x 5,0 y 5 2 , 0 xy 25 2(x+1)(2y+1) xy (2xy+x+2y+1) xy 2 xy+(x+2y+1) xy 2 xy+6 xy 5 2+3 5 and xy minimum infinity tends to 0, so the minimum value of the original formula is 6. Hope you can close the town to help you! Thank you!

Solution: Let baim=1 2 x=2 (-x), which is the subtraction function on the domain du.

Then: zhix [-3,dao2], m [1 4,8]. The special quadratic function f(x)=g(m)=m 2-m+1=(m-1 2) 2+3 4, when m=1 2, the function has a minimum value: g(m)=3 4, where x=1;

When m=8, the function has a maximum value: g(m)=57, where x=-3.

So the maximum and minimum values of the function f(x)=1 4 x-1 2 x+1 are: 57, 3 4, respectively.

Answer:

f(x)=1/4^x-1/2^x+1

1/2^x-1/2)^2+3/4

df/dx=2(1/2^x-1/2)(-1/4^x)*2^x*ln2=0

When 1 2 x-1 2 = 0, x = 1, f(x) min = 3 4

1 4 x = 0 and 2 x * = 0 are not true.

That is, the function f(x)=1 4 x-1 2 x+1 has only one inflection point.

In defining an extreme value within the domain, the back can only be answered from its increment or subtraction.

Sexual investigations. When x<1, f(x)=1 4 x-1 2 x+1 is a decreasing function, so it must have an extreme value when x=-3, f(-3)max1=57, when x>1, f(x)=1 4 x-1 2 x+1 is an increasing function, so it must have an extreme value when x=2, f(2)max2=13 16, and the above results show that f(x)=1 4 x-1 2 x+1 has a maximum value of 57 and a minimum value of 3 4 in x [-3,2].

f(x)=(1 4) copy x-(1 2) x+1 set bai(1 2) x=t 0

f(x)=t 2-t+1=(t-1 2) 2+(3 4) when x du[-3,2], the maximum value.

zhi:daox=-3,t=8 f(x)=73 min:t=1 2 f(x)=3 4

Element 1 2 x=t x "-3,2" Exponential function monotonicity t "1 4,8".

f(x)=g(t)=t2-t+1 The minimum value of the monotonicity of the number in the quadratic function fmin=g(capacity 1 2)=3 4

The maximum value fmax = g(8) = 57

y=-2x²+x=-2(x²-x/2+1/16)+1/8=-2(x-1/4)²+1/8

So when x = 1 4, the maximum value of y is 1 8

Let t = root number 1-6x

t 0) then t = 1-6x

x=1-t²/6

Original formula = 2 (1-t 6) + t

1-t²+3t/3

t 0) The following is the method of the conventional evaluation range to find the maximum value.

Solution: There is only one element in a {x r ax +2x+1=0, a r}.

Equation. ax buried retardation +2x+1=0

The two equal real numbers are bent and voided.

2^2-4a=0

The solution is a=1, and a=1 is substituted into the danxiao ax +2x+1=0.

The solution is x=-1

Therefore, this element in a=1a is.

1 is a=

A {x r ax +2x+1=0,a r} has only one element.

If Gao Chawei a=0, 2x+1=0, x=-1 2 if a≠o, that is, the equation.

ax²+2x+1=0

The two are equal.

Therefore, the discriminant formula is 0

i.e. 2 2-4a*1=0

So no game. a=1, substituting the original Qi Pei equation to obtain x=-1

A {x r ax +2x+1=0, a r} has only one element in the grinding hand.

i.e. equations. ax²+2x+1=0

The two are equal, so the discriminant formula is 0:

2^2-4a*1=0

So. a=1, the original equation is.

x²+2x+1=0

Blind to the solution. x=-1, therefore. a=.

Find out the median of these endpoints.

1 1, 2 1 2, 3 1 3,,, 2011 1 2011 Find out the median, 1 1422

So, the minimum value f(1 1422).

f(x) x socks - 2ax + 1

X -2ax + A mountain + 1 - A (X - A) 1 - A

When a < 1, the minimum value = f(1) =2a + 2 when a > 3, the minimum value = f(3) =6a + 10 when -1 a 3 is funny, the minimum value = f(a) =1 - a

Let's talk about the axis of symmetry.