Given that the maximum value of the function y x2 2ax 1 on the closed interval 1,2 is 4, then the va

The axis of symmetry of the function is x=-a

When -a -1 => a 1 (i.e., x is taken to the right of the axis of symmetry), then x=2,y is taken to the maximum, i.e., 2 +2a*2+1=4 => a=-1 4 (rounded).

When -a 2 => a -2 (i.e., x takes the value to the left of the axis of symmetry), then x=-1, y takes the maximum value, i.e., (

1) +2a*(-1)+1=4 => a=-1 (rounded).

When -1 -a 2 = > -2 a 1 (i.e., the symmetry axis is within the range of x), (a)-(1)<2-(-a) = > a>-1 2 (i.e., the symmetrical wheelbase x=-1 is closer), then x=2,y is taken to the maximum, a=-1 4

a)-(1)>2-(-a) => a<-1 2 (i.e., the symmetrical wheelbase x=2 is closer), at this time x=-1, y takes the maximum value, a=-1

a)-(1)=2-(-a) => a=-1 2 (i.e., the axis of symmetry and x=-1 are equidistant from x=2), in this case, x=-1 and a=-1 2 are brought into the function, and y=3≠4 (rounded).

In summary, a = -1 4 or -1

Draw a picture yourself to understand.

y=x2+2ax+1=4 y=x2+2ax3

Since the maximum value on the closed interval [-1,2] is 4, y=(x+1)(x-2)=x2-3x-3<=o y is the maximum.

So, a=-3 2

The axis of symmetry of the function is x=-a

When -a -1 => a 1 (i.e., x is taken to the right of the axis of symmetry), then x=2,y is taken to the maximum, i.e., 2 +2a*2+1=4 => a=-1 4 (rounded).

When -a 2 => a -2 (i.e., x takes the value to the left of the axis of symmetry), then x=-1, y takes the maximum value, i.e., (

1) +2a*(-1)+1=4 => a=-1 (rounded).

When -1 -a 2 = > -2 a 1 (i.e., the symmetry axis is within the range of x), (a)-(1)<2-(-a) = > a>-1 2 (i.e., the symmetrical wheelbase x=-1 is closer), then x=2,y is taken to the maximum, a=-1 4

a)-(1)>2-(-a) => a<-1 2 (i.e., the symmetrical wheelbase x=2 is closer), at this time x=-1, y takes the maximum value, a=-1

a)-(1)=2-(-a) => a=-1 2 (i.e., the axis of symmetry and x=-1 are equidistant from x=2), in this case, x=-1 and a=-1 2 are brought into the function, and y=3≠4 (rounded).

In summary, a = -1 4 or -1

Draw a picture yourself to understand.

It's best to draw a picture, it's very simple, decide it yourself

Discussed in case scenario, when the axis of symmetry is to the left of -1 or the axis of symmetry is to the left of -1 to 2 or the axis of symmetry is to the right of 2. Careful calculation can be made.

Solution: Let t=a x, then: y=t 2+2t-1, 1 a<=t<=a, t(a>1) is the increasing function.

The function y increases monotonically over the interval [-1,1], so: t=aa 2+2a-1=14

Solution: a=3, a=-5<1 (rounded).

A 2x and a x are both functions of a century in the range of -1 to 1, so when x = 1 is maximum, a 2 + 2 a-1 = 14 gives a = 3 or -5, because a > 1, so, a = 3

Solution: y=a 2x+2a x-1=(a x+1) 2-2,(-1 x 1).

Let t=a 2x, then f(t)=(t+1) 2-2(1) when the 0 function f(t)=(t+1) 2-2 is an increasing function on [a,1 a] f(t)max=f(1 a)=[(1 a)+1] 2-2=4, i.e. [(1 a)+1] 2=6

Solution: a=(1+ 6) 5 or a=(1- 6) 5 (rounded).

-x2+ax-a/4+1/2=-(x-a/2)²+a²/4-a/4+1/2

It is known that the maximum value of the function y=-x2+ax-a 4+1 2 on the interval [0,1] is 2, so there is.

1) A2 [0,1], A4-A4+1 2=2 does not meet the requirements of A

2)a/2>1,f(1)=-1+a-a/4+1/2=2a=10/3

3)a/2<0,f(0)=-a/4+1/2=2a=-6

Synthesis (1).(2).(3) The value of the real number a is -6 or 10 3

Let t=a x

then y=t 2+2t-1=(t+1) 2-2, when y takes the maximum value, i.e.

t+1)^2-2=14

t1=-5 and t2=3, i.e.

a x=-5 (contradicts a 0, rounded).

or a x = 3;Because.

a x in. -1,1] is a monotonic function, so you get.

a^(-1)=3

Get a = 1 3

or a 1=3

Get a=3 and get it.

a=1 3 or a=3

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.

If the function y=ax 2=2ax(a≠0) has a maximum value of 3 in the interval [0,3], find the value of a.

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

When a<0, the function obtains the maximum value -a=3 at the vertex x=1, and obtains a=-3;

When a>0, the function obtains the maximum value f(3)=9a-6a=3 at x=3 to obtain: a=1

In summary: a = -3 or 1

You don't have to make a drawing.

Suppose f(x)=ax 2-2ax

The symmetry axis of the quadratic function is x=1, over the interval [0,3].

When a>0 and the function opening is upward, then the maximum value is f(3) When a<0 and the function opening is downward, then the maximum value is f(1) so as long as a≠0, the function always has a maximum value in the interval.

Let m=a baix

y=m²+2m-1

m+1)²-2

The opening is upward, and the axis of symmetry m=-1

Because dum=a x>0>-1

So zhim +2m-1 increments.

then m is the largest when daoy=14.

So 0, then x=-1, max=1 a +2 a-1=14, a=1 3 the same way, a>1, then a +2a-1=14, a=3 a=1 3, a=3

Solution: y=-x +ax-a 4+1 2 is a parabola with an opening downward and its axis of symmetry is x=a 2, when 0 a 2 1, i.e., 0 a 2, x = a 2, y is the maximum.

a 4 + a 2 - a 4 + 1 2 = 2 a -a - 6 = 0, a = -2 or a = 3 is not suitable (0 a 2).When x=a2 0i.e. a 0,x [0,1].

y is a subtraction function. x=0, and y takes the maximum value as 2Namely:

a 4 + 1 2 = suitable for a 0) x = a 2 1, that is, a 2, y is an increasing function, x = 1, y takes the maximum value as 2i.e.: -1+a-a 4+1 2=suitable for a 2) The value of the real number a is:

a=-6 or a=10 3