f x x 2 3x 2 x 2 2x 1 to find the minimum of the function

y=(x 2-3x+2) (x 2+2x+1), define the field: x is not equal to -1 move and tidy:

x 2(y-1)+x(2y+3)+y-2=0 discussion: when y=1; x exists, so y can be 1

When y is not equal to 1, x exists, so the discriminant of the quadratic equation about x" = 0, i.e., > = 0

So (2y+3) 2-4(y-1)(y-2)>=024y+1>=0

So y>=-1 24

So the minimum of f(x) is -1 24

Lingy'>0 gives x (negative infinity, 0) (1, positive infinity) so f(x) increases monotonically in (negative infinity, 0) (1, positive infinity) and decreases monotonically in (0, 1).

So take the maximum value f(0)=0 at x=0

Take the minimum value f(1)=2-3=-1 at x=1

x [ 2 2, 3 2], according to the basic inequality.

f(x)＝x/(x^2+3)

x/(2√3x)

x 2 shed section base 3, i.e. x 3, f(x)|max＝√3/6.

At the same time, f(x)|min＝min

f(3 2) or f( chain 2 2).

f(x)=2x (x+1) is obtained from f(x)=2-2 Wu Chun (x+1) is the function of f(x)=-2 x is the function of 2 units upward and 1 unit to the left f(x) is the increasing function on (- 1) and (-1, +) [-3,-2] is included in (- 1) mountain shouts f(x)max=f(-2)=4f(x)min=f(-3)=3 Answer: The maximum value of the function.

Summary. First, let's find the function $f(x)=x 2-2|x|+3$ minimum. Since $f(x)$ is a quadratic function, the minimum value can be found based on its axis of symmetry position.

When $x geq0$, $f(x)=x 2-2x+3$; When $x 0$ and $6$ unequal real roots. The solution is $d=(a-4) 2>16$, which is $a 12$. And because it's $d>0$, it's $aeq 4$.

Therefore, the range of $a$ is $a 12$ and $aeq 4$.

315.The pure scatter f(x)=x 2-2|x|+3 with a minimum value; If the equation f 2(x)-(a+2)f(x)+2a=0 about x has exactly 6 unequal real root land shots, then we can find the value range of a.

No problem. First, let's find the function $f(x)=x 2-2|x|+3$ minimum. Since $f(x)$ is a quadratic function, the minimum value can be obtained by the osmosis based on its symmetry axis position.

When $x geq0$, $f(x)=x 2-2x+3$; When $x 0$ and there is a pat core of $6$, unequal real roots. The solution is $d=(a-4) 2>16$, which is $a 12$. And because it's $d>0$, it's $aeq 4$.

Therefore, the range of $a$ is $a 12$ and $aeq 4$.

Is a less than minus four or a greater than 12 pairs?

That's right. Appreciate.

No problem. It is inconvenient to ask another question.

You said. This doesn't seem to be able to do it for you, because I can't see ** here. I can't do it, baby.

f(x)=2(丨x-1 2丨virtual slip+丨x-3 2丨)>=2|x-1/2-x+3/2|=2

The minimum difference is 2

At this point, x is at [1 2, 3 2].

My rotten socks are the most comprehensive,

Because f(x)=x2+2x+3

f(x)=x^2+2x+1+2

f(x, so the value of f(x) should be the smallest, but also the value of f(x) 0, that is, x 2+2x+3 0, that is, (x+1) 2+2 0

The slag is excavated with (x+1) feet quietly 2+2 0

x+1）^2≥-2

And (x+1)2 has a minimum value of 0, so.

When x=-1, the function f(x)=x 2+2x+3 has the smallest value!

Derivative f'(x)=1-3/2*2/3*x^(2/3-1)=1-x^(-1/3)

So -1 x 0 and 1 x 27 8 o'clock'(x) 0, which decrements when the function is incremented by 0 x 1.

So the minimum value may be taken at -1 or 1.

f(-1)=-5/2，f(1)=-1/2

So the minimum value is -5 2

f(x)=(x 2+2x+3) x=x+2+3 x,x [2, let 2 x10

x1x2 4, 1 4 (1-3 x1x2)< 1 (x2-x1)(1-3 x1x2)>0 i.e. f(x2)-f(x1)>0 f(x2)>f(x1).

f(x) is an increment function on [2,+.

When x = 2, the minimum value of f(x) = f(2) = 11 2

Note: If you use the basic inequality x+3 x 2 3, it is not correct, because the condition for taking the equal sign is x=3 x, that is, x= 3, 3 is not in [2,+, so the equal sign cannot be taken.