Mathematics knows that the function f x 4 x a2 x b, when x 1, f x has a minimum value of 1

Let 2 x= t f(t)=t 2 - a·t +b is a quadratic function.

When t=a2, f(t) has a minimum value, i.e., f(x) has a minimum value—1 (when x=1).

In this case, t = 2 x =2 so a=4 is substituted to obtain b=3f(t)= t 2 - 4t +3 0 solution to get 1 t 3 i.e. 1 2 x 3 solution to log2 1 x log2 3 i.e. 0 x log2 3

Mathematics: The function f(x)=4 x-a2 x+b is known, and when x=1, f(x) has a minimum value of -1 to know.

Solution: (1) Let 2 x=t, then f(t)=t 2-at+b, when x=1, that is, t=2, f(x) has a minimum value of -1. Since f(t) is a quadratic function, when t=a2=2, f(x) has a minimum value of —1, so a=4.

When x=1, 4-2a+b=-1, a=4, so b=3

2) f(t)=t 2-4t+3 0, 1 t 3, so x [0, log23] (is the logarithm of 2 base 3, not 23), that is, the set a = [0, log23].

Let g(x)=f(x)-(1 2x 2+ax+b)=e x-(a+1)x-b>=0

then g'(x)=e^x-a-1

For example, if the orange fruit is pie, Sun Hui will be envious of Sun Hui -a-1>=0, that is, a

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

If and only if x-1=1 (x-1), i.e., the source is cleared x=2, there is a small value of f(x)min=4

f(x)=4^x-a2^(x+1)+b=f(x)=2^（2x）-2a（2^x）+b

Another 2 x=t, so t>0

Then f(t) concession = t 2-2at+b=(t-a) 2+b-a 2, when x=2, f(x) has a minimum value of 10, then the quiet town game, that is, t=4, f(t) minimum value is 10

So b-a 2 = 10 and the axis of symmetry t = a = 4

The solution is that the travel consumption is a=4 and b=26

a^2+b^2>=2ab

f(0)=a+2b=4

a+2b)^2=16

a^2+4b^2+4ab=16

4ab+4ab<=16

ab<=2

Substituting x=1,a+2b=4 into f(x)=x 2+abx+a+2b gives f(1)=1+ab+4<=7

That is, the maximum value of f(1) is 7

It is known that f is obtained'Volt sell(x)=e x(ax 2+2ax+a+1) when a=0, f'(x)=e x>0 At this point f(x) is monotonically increasing, so at x [-2,-1 ], f(x) f(-2)=e (-2) contradicts f(x) 2 e 2 with the lack of laughter, so a=0 is not eligible, so a≠0 when a>0, and the discriminant of ax 2+2ax+a+1 = (2a) 2-4a(a+1)=-4a<0 So for any x, ax 2+2ax+a+1>0, so f'(x) >0, i.e. f(x) increases monotonically on [-2,-1], so f(x) f(-2)=(4a+1+1) e 2=(5a+1) e 2 e 2 so 5a+1 2, thus a 1 5When a<0, the solution of ax 2+2ax+a+1=a(x+1) 2+1=0 is x=sqrt(-1 a)+1, or x=-sqrt(-1 a)-1In the discussion interval can only be x=-sqrt(-1 a)-1

1) If L-Qin x=-sqrt(-1 a)-1 -2, i.e. -1 a<0, f'(x) is greater than 0 on [-2,-1], so f(x) increases monotonically at [-2,-1], so f(x) f(-2)=(5a+1) e 2 2 e 2, the solution a 1 5 does not conform to 2) if x=-sqrt(-1 a)-1>-2, i.e. a<-1 when f'(x) is less than 0 on [-2,x] and f on [x,-1].'(x)>0, so f(x) decreases monotonically on [-2,x] and increases monotonically on [x,-1], so f(x) f(-sqrt(-1 a)-1)=2a(sqrt(-1 a)+1)e (-sqrt(-1 a)-1), while f(-sqrt(-1 a)-1)<0, so it cannot be greater than or equal to 2 e 2, so a<-1 does not match. In summary, the value range of a is a>=1 5

Let a be a real number and the function f(x)=x 2+|x-a|+1,x r, find the minimum value of f(x).

i'When x a, f(x)=x 2+x-a+1=[x 2+x+(1 4)]-a+1-(1 4).

x+(1/2)]^2+(3/4)-a

Its axis of symmetry is x=-1 2

Then, when a -1 2, because of x a, then the function f(x) can obtain the minimum value f(-1 2) = (3 4)-a;

When a>-1 2, because x a, because the opening is upward, then the minimum value of the function f(x) is f(a)=a2+1;

ii'When x a, f(x)=x 2+x-a+1=[x 2+x+(1 4)]-a+1-(1 4).

x-(1/2)]^2+(3/4)+a

Its axis of symmetry is x=-1 2

Then, when a -1 2, because of x a, then the function f(x) can obtain the minimum value f(-1 2) = (3 4)-a;

When a>-1 2, because x a, because the opening is upward, then the minimum value of the function f(x) is f(a)=a2+1;

In summary: when a -1 2, the function f(x) has a minimum value (3 4)-a;

When -1 2 is a 1 2, the function f(x) has a minimum value of (3 4) + a.

As for when the equal sign is taken, because at a=-1 2 or a=1 2, both endpoint values are equal. That is, the final expression of the function with respect to the minimum value of f(x) is a continuous function. ）

From f(0)=4, we can see that a+2b=4

Then find the maximum value of ab according to the mean value theorem.

a+2b>=2*root number (a2b).

4>=2*Root number (A2B).

2> = root number (a2b).

4>=2ab

2>=ab

So the maximum value of ab is 2

f(1)=1+2+4=7

Solution: This question examines the monotonicity of composite functions: for the monotonicity of conforming functions, remember the phrase "the same increases and decreases"; This means that there is such a composite function y=f(x)*g(x); When f(x) and g(x) have the same monotonicity, the function y is a monotonic increasing function.

Back to this topic: f(x)=(2ax+a+1)e x ;

1) When a!=0: We can see f(x)=g(x)*m(x): g(x)=-2ax+a+1, m(x)=e x

According to the properties of the exponential function and the primary function, we can know that m(x) is a monotonic increasing function in r, and when 02) when a=0 f(x)=e x is very obvious to Li Mingshi f(x) is a monotonic increasing function on r, then in the given interval [0,1], when x=1 obtains the maximum value f(1)=e

It is composed of a composite function, one is a one-time function y1=-2ax+a+1, the other is an exponential function y2=e x, to find the maximum and minimum value to see the monotonicity, the composite function to see the monotonicity of the two sub-functions, where y1 is the subtraction function, y2 is the increase function, the mu is the subtraction function, there is a problem, that is, a, a=0, f(x) is equal to e x, then it is simple f(1) is the largest, f(0) is the smallest; a is not equal to 0That is, f(0) is the largest and f(1) is the smallest.