Find the minimum value of the function f x x 2 ax 4 over the interval 1,2

Axis of symmetry: x=-a 2

When -a 2<=1, that is, the function a>=-2 increases monotonically on [1,2], then the minimum value is taken when x=1, and ymin=5+a

When -a 2>=2, i.e., a<=-4, the function decreases monotonically on [1,2], and the minimum value is taken when x=2, and ymix=8+2a

When -4

Recipe: f(x)=(x+a2)2+4-a2 4

Discussion: Divide the number axis into three segments with 1 and 2, discuss the increase and decrease of f(x) at -a 2 in these three intervals, and find the extreme value.

Solution: f(x)=x 2+ax+4=(x+a 2) 2+(16-a 2) 4

When -a 2<1 is a>-2, f(x) is in the interval [1,2] f(x)>=f(1)=5+a

In this case, f(x) takes the minimum value of 5+a at x=1

When 1<=-a2<=2, i.e., -4<=a<=-2, f(x) obtains a minimum value in the interval [1,2] as f(-a 2)=(16-a2) 4

When -a 2>2 is a<-4, f(x) obtains a minimum value in the interval [1,2] as f(2)=8+2a

When the above a>-2 is combined, the minimum value of f(x) is 5+a

4<=a<=-2, the minimum value of f(x) is (16-a2) 4

The minimum value of f(x) at a<-4 is f(2)=8+2a

When the number source A is greater than -2 and less than or equal to 1.

The minimum is f(-2) and the maximum is f(a).

When a is greater than one and less than or equal to 4.

The minimum is f(-2) and the maximum is f(1).

When a is greater than 4, the cautious Bi faction.

The minimum is wide f(a) and the maximum is f(1).

f(x)=x^2-2ax-1=（x-a）^2-1-a^2;When a<0, the function increases monotonically over the interval [0,2], with the minimum value of f(0)=-1 and the maximum value of f(2)=3-4a. When 0<=a<=1, the function decreases monotonically in the interval [0,a] and increases monotonically in the interval (a,2), the minimum value is f(a)=-1-a 2, and the maximum value is f(2)=3-4a; When 1 asks God to leak! When < a<=2, the function decreases monotonically over the interval [0,a] and increases monotonically on the interval (a,2), with the minimum value of f(a)=-1-a 2 and the maximum value of f(0)=-1;When 2

f(x)=(x-a)^2-1-a^2

The opening is upward, the axis of symmetry is x=a, according to the position relationship between the axis of symmetry and the interval, we get:

a<0, fmin=f(0)=-1, fmax=f(2)=3-4a0=2, fmin=f(2)=3-4a, fmax=f(0)=-1

As can be seen from the question, the axis of symmetry of f(x) is x a

1°, when a 0 and f(x) increases monotonically at 0 2, then the maximum value f(x) f(2) 4 4a 1, and the minimum value f(x) f(0) 1

2°, when a 2, f(x) decreases monotonically at 0,2, then the maximum value f(x) f(0) 1, the maximum value f(x) f(2) 4 4a 1

3°, when 0 a 1, the minimum value is obtained at the axis of symmetry, f(x) f(a) a 2 1, the maximum value is obtained at x 2, and the minimum value is f(x) f(2) 4 4a 1

4°, when 1 a 2, the minimum value is still obtained at the axis of symmetry, the value is a 2 1, the maximum value is obtained at 0, and the maximum value f(x) f(0) 1

f（x）=2x^2-2ax+3

2(x^2-ax+a^2/4)+3-a^2/2=2(x-a/2)^2+3-a^2/2

When a 2<-1 i.e. a<-2 i.e., the parabolic vertex is at the left point of x=-1, then the minimum value is f(-1)=2*(-1) 2-2*a*(-1)+3=2+2a+3

5+2a, when -11, i.e., a>2, i.e., when the parabolic vertex is to the right of x=1, then the minimum value f(1)=2*1, 2-2a*1+3=5-2a

f(x)=x^2-2ax+4

x-a) Rubber roller +4-a

0 minimum = 4

The minimum value of locust = 16-8a + 4 = 20-8a

Answer: Prove that f(x) is a subtraction function on (0,2) and an increasing function on (2,+), either x1 and x2 belong to (0,2).

Let x10 and 02

f(x) is a subtraction function at [1,2] and an increasing function at [2,a] The minimum value is f(2)=2+4 2=4

Have you ever learned derivatives ?

Analysis: With the help of a +b 2ab, find the trough point, i.e., x+4 x 2 x*4 x 4, when x=4 x, i.e., x=2, where f(x) is the smallest.

So the solution: when a 2.

Using the definition of the monotonicity of the function, it is proved that f(x) is decreasing on [1,a], so when x=a, the minimum value is a+4 a, and when a 2 is a , the monotonicity definition of the function can be used to prove that f(x) decreases on [1,2] and increases monotonically on 2,a.

So when x=2, the minimum value is obtained as 4].

x^2+2ax+1=（x+a）^2 + 1-a^2

The minimum value of f(x) is f(x)=1-a2 when x=-a

1) When -1<=-a<=2, i.e. -2<=a<=1, the minimum value of f(x) in the interval [-1,2] is 1-a 2=-4 so a 2=5

a=root5 or a=- root5 but does not meet the condition of -1<=a<=2, so the assumption is not valid.

2) When -a<-1, i.e., a>1, f(x) increases monotonically in the interval [-1,2], so the minimum value of f(x) is f(-1)=1-2a+1=2-2a=-4

So a=3

3) When -a>2, i.e., a<-2, f(x) decreases monotonically in the interval [-1,2], so the minimum value of f(x) is f(2)=4+4a+1=-4

So 4a=-9 a=-9 4

In summary, a=3 or a=-9 4

If a=1

then the minimum is -1-a 2

The maximum is f(0)=-1

If 0=2, the maximum is f(0)=-1

Minimum f(2) = 3-4a

If a<=0

The maximum is f(2)=3-4a

The minimum is f(0)=-1

Derivative, which has a derivative of 2x-2a

Let it be equal to 0, then x=a

When x=a, f(x)=-a 2-1 This is the maximum value x=0, and f(x)=-1

When x=2, f(x)=3-4a

Discuss the magnitude of the value at x=0,2, the smaller of the two is the minimum value of this interval.

Answer: The minimum value between [0,2] is: -1 if a<=0;-a 2-1 if 0=2;

The maximum value is: -4a+3 if a<=1; -1 if a>1

The main observation is that this is an upward parabola, and the axis of symmetry is x=a, so the above result can be easily obtained.