The known function f x x 2 2x 3 1 discusses the minimum value g t on the interval t 1, t

1）f(x)=x²+2x-3=(x+1)²-4

Axis of symmetry: x=-1, opening up.

t-1,t] on the left side of the axis of symmetry, i.e., t-1, f(x) decreases monotonically, g(t) = t +2t-3

t-1,t] contains the axis of symmetry, i.e., ->-1t 0 at -1-1, f(x) monotonically increasing, g(t) = (t-1) +2(t-1)-3=t -4

2) [m,m+2] to the left of the axis of symmetry: max=f(m)=m +2m-3=5 m=-4, (m=2>-1, rounded).

m,m+2] on the right side of the axis of symmetry: maximum = f(m) = (m+2) +2(m+2)-3 = 5

m=0, (m=-6<-1, rounded).

The value range of m is -4,0

f(x)=(x+1) 2-4 decreases monotonically at x<-1 and increases monotonically at x>-1.

The scope of t is discussed next.

When t-1<=-1<=t, i.e., -1<=t<=-2, g(t)=-4, when t<=-1, g(t)=f(t)=t 2+2t-3, when t-1>=-1, i.e., t>0, g(t)=f(t-1)=t 2-4 second sub-question. Due to the monotonicity of the function, the maximum value of the function must occur at x=m or x=m+2.

Let f(m)=5 and f(m+2)=5 respectively, and m2=9 and (m+2) 2=9, respectively

Take m=3, m+2=5, and the maximum value of the 5 function is 21, which is excluded.

Take m=-3 and m+2=-1 to meet the conditions.

Take m=1, m+2=3, and meet the conditions.

Take m=-5, m+2=-3, the maximum value of the -5 function, the maximum value is 21, and exclude.

So m=1 or m=-3

f(x)=x 2-2x+2=(x-1) 2+1, when x=1, f(x) has a minimum value of f(1)=1

The next thing is to take advantage of the monotonicity of the quadratic function).

f(x) decreases monotonically on (- 1) and increases monotonically on stove bend (1,+, so.

When the x=1 bit is argued to the right of t, t+1 is t+1

From the question we get f(x) (x-1) 2-2

is a parabola with an opening of (1,-2) vertices pointing upward.

g(t) f(t+1) when t+1 is less than or equal to 1, and g(t) f(t) when t>=1

When t<1=1 g(t) t 2-2t-1

1) f(x)=-x +4x-1=-(x-2) +3, f(x) is a quadratic function of the old book in x, the opening is downward, and the axis of symmetry x=2

If 2 (t,t+1), i.e., 1<=t<=2, the maximum value of f(x) is obtained at x=2.

g(t)=f(2)=3

If t>=2, then f(x) decreases monotonically at [t,t+1], and the maximum value of f(x) is obtained at x=t.

g(t)=f(t)=-t-2)²+3

If t<=1, f(x) increases monotonically on [t,t+1], and the maximum value of f(x) is obtained at x=t+1.

g(t)=f(t+1)=-t+1-2)²+3=-(t-1)²+3

g(t) is analytically .

When t>=2, g(t)=-t-2) +3, g(t) achieves a maximum value of 3 at t=2

When 1t<=1, g(t)=-t-1) +3, and g(t) achieves a maximum value of 3 at x=1

In summary, the maximum value of g(t) is 3

From the question we get f(x) (x-1) 2-2

is a parabola with an opening of (1,-2) vertices pointing upward.

g(t) f(t+1) when t+1 is less than or equal to 1, and g(t) f(t) when t>=1

When t<1=1 g(t) t 2-2t-1

From the problem, it can be obtained: f(x) (x-1) 2-2 is a parabola with an opening upward vertex of (1,-2), g(t) f(t+1) when t+1 is less than or equal to 1, g(t) f(t) when t>=1, g(t) t(t) 2-2t-1 when t<1=1

Shou Wu Kai transforms first:

f(x)=-x^2-2tx)+3

x^2-2tx+t^2-t^2)+3

x-t)^2+3+t^2

Then categorize and discuss:

1<=t<=1: when x=t, take the maximum value of the large object t 2+3t>1: when x=1, take the cavity and imitate the maximum value -(1-t) 2+3+t 2=2+2t

t<1: When x=-1, the maximum value is 2-2t

Hope it works.

f(x)=-x-2tx+t)+t+t+3-(x-t)squared dismantling the ant+t +3

It can be seen as burying a parabola to the brigade and finding the vertex value.

1.When t -1: Maximum:

x=-1，f(x)=2-2t2.When t -1: Maximum:

x=1，f(x)=4+2t3.When -1 "this hail = t "=1: maximum, f(x)=t +3