It is known that the function y 2ax 1 x 2 finds the maximum value of y over the interval 0,1

Analysis: To solve the maximum-value problem, it is generally necessary to determine the monotonicity of the function in the interval.

This question contains the parameter a, so it is necessary to discuss a to obtain the corresponding monotonicity.

Solution: y'=2a+2/x^3=2/x^3(ax^3+1)a≥0，y'> 0, so y increases monotonically over the interval (0,1), ymax=y(1)=2a-1

a<0,y'=2a x 3(x 3-(-1 a))1)-1 a 1, i.e. -1 a<0,y'0,y increases monotonically in the interval (0,1), ymax=y(1)=2a-1

2)-1 a<1, i.e., a<-1 so y increases at (0, under the third root sign (-1 a)) and decreases up (under the third root number (-1 a), 1], so ymax=-3(-1 a) (2 3).

In summary: when a -1, ymax=2a-1, when a<-1, ymax=-3(-1 a) (2 3) hope it can help you!

At x10, y1-y2=(x1-x2)(2a+(x1+x2) (x1 2*x2 2)<0, and the function y increases monotonically.

y is at the maximum value of the interval (0,1), x=1;y=2a-1;

When a=0, y is the maximum value on the interval (0,1), x=1;y=-1;

When a<0; 2a+(x1+x2) (x1*x2) 2>0, y1-y2<0, function y monotonically incremental, y maximum, y=2a-1

When a<0; When 2a+(x1+x2) (x1*x2) 2<0, y1-y2>0, the function decreases monotonically, and there is no maximum value for y.

Solution: Because 0 1 2 1, so function.

y (1 2) is a decreasing function of the single split hall on the interval [2,3].

When the source is not x 2, take the maximum value ymax (1 2) 2 1 4;

When x 3, take the minimum value ymin (1 2) 3 1 8.

y=x^2+4x-1

x^2+4x+4-5

x+2)^2-5

The axis of symmetry x=-2, the opening is upward, and the given interval [-1,1] is on the right side of the axis of symmetry, so it is a monotonically increasing interval, so:

ymax=f(1)=3^2-5=4;

ymin=f(-1)=1^2-5=-4.

y=a^2x+2a^x-1=(a^x)^2+2a^x+1-2[a^x+1]^2-2

Let a x=t 0, then:

y=(t+1) 2-2 represents a quadratic function with t=-1 as the axis of symmetry and the opening upwards: when 0 a 1, a x is a subtraction function, and the maximum value of t=a x on [-1,1] is t=a -1=1 a 1

In this case, the maximum value of y is y=[(1 a)+1] 2-2=14, then a=1 3

When a 1, a x is the increasing function, and the maximum value of t=a x on [-1,1] is t=a 1=a 1

In this case, the maximum value of y is y=[a+1] 2-2=14, then a=3

In summary, a = 1 3 or a = 3

Solution: Let t=a x, then y=t 2+2t-1

The axis of symmetry of the function is t=-1

When a (0,1), t [a,1 a].

x=-1 to the left of (a,1 a).

y(t)max=y(1 a)=14,a1=1 3,a2=-1 5(rounding).

When a (1,+, t [1 a,a].

Similarly, x=-1 is to the left of (1 a,a).

y(t)max=y(a)=14,a1=3,a2=-5(round)In summary: a=1 3 or 3

I analyze and analyze...

Look at the axis of symmetry, there are three situations:

1, a 2<0, then the maximum value appears in 0 or 1, and the value of 2 a is substituted into the solution. However, as long as the selection is calculated by 0 or 1, it means that the monotonicity is determined, and after solving a, the analytical formula of this situation must be expressed, and then the monotonicity is judged to be in line with this situation;

, the same is true in the first case, but it also has to be verified separately depending on the situation.

As for where 0 and 1 go, I'll see for myself.,I'll talk about my thoughts.。。 It's not even free, it's too late to go to bed!!

Hey, hey, it's good to help others, comment on the answers, chase the score!!

y=x^2+2ax+1

x^2+2ax+a^2-a^2+1

x+a)^2+1-a^2

The axis of symmetry is x=-a

When -a=1 2: f(x)max=f(2)=f(-1)=1-2a+1=4, a=-1 (contradictory to a=-1 2), rounded.

When -a [-1,1 2): f(x)max=f(2)=4+4a+1=4,-a=1 4 [-1,1 2),a=-1 4

When -a (1 2,2]: f(x)max=f(-1)=1-2a+1=4,-a=1 (1 2,2],a=-1

When -a (-1): f(x)max=f(2)=4+4a+1=4, -a=1 4 (contradictory to -a (-1)), rounded.

When -a (2,+): f(x)max=f(-1)=1-2a+1=4, -a=1 (contradictory to -a(2,+), rounded.

In summary: a=-1 4 or a=-1

Let t=a x, then the function is y=a 2x+2a x-1=t 2+2t-1=(t+1) 2-2 From the equation y=t 2+2t-1, we can see that the curve is open upward, and the parabola of the point (0,-1) is obviously y=t=a x>0, so the original function y=a 2x+2a x-1 curve is on the right side of the y-axis (increasing function); In order to obtain the maximum value of the original function y=a 2x+2a x-1, it is necessary to take the maximum value of y=t=a x, and in the interval [-1,1], when a > 1, y=t=a x (the increasing function) has a maximum value a; (1) When 0< a<1, y=t=ax (subtraction function) has a maximum value of 1 a(2) Substituting y=t=a or 1 a into (t+1) 2-2=14 respectively can obtain a=3 or a=1 3.

y=(x+a)^2+1-a^2

The maximum value must be obtained at the endpoint:

y(-1)=-2a

y(2)=5+4a

If y(-1)>y(2), i.e. a<-5 6, then the maximum value = y(-1)=-2a=4-->a=-2

If y(-1)<=y(2), i.e., a>=-5 6, then the maximum value = y(2)=5+4a=4-->a=-1 4

Solution: y=a 2x+(2a x)-1=y=(a x+1) 2-2; Because when a>1, a x >0 and a x is an increasing function, there is a x>=a in the interval [-1,1], so that ymax=(a+1) 2-2=14; So there is a+1=4; thus a=3;