If the range of the function y fx is 0 5 to 3 then the range of the function fx 1 fx is the process

Solution: f(x) = f(x) +1 f(x).

If f(x) = t, then f(t) = t + 1 t ( t 3);

In this case, the original problem is transformed into the range of f(t) = t + 1 t when t [,3].

For f(t) = t + 1 t we can find its monotonic interval by finding the first derivative:

Since f(t)= 1 - 1 t 2, by making it greater than 0 we can get when |t|1, f(t) 0, which is also a monotonically increasing interval of f(t) = t + 1 t, the same should be |t|<1, and t≠0, f(t) is monotonically decreasing (which I don't quite say), so when t 3, a decreasing interval [,1] and an increasing interval [1,3] are crossed.

Apparently f(t) achieves a minimum value at t=1, i.e., f(t)= 2;

Decreasing on [,1] and increasing on [1,3], so when determining the maximum, we only need to compare the magnitudes of f( and f(3), obviously f(3) = 10 3 max , so the range of the original problem is [2,10 3]!

Here's a sketch I drew for you to see.

By the way, I'd like to add! You may wonder why I determined that the range of values is between these two numbers after determining the minimum and maximum values. In fact, it depends on the continuity of the function, to put it simply, that is, f(t) = t + 1 t When t [,3], it is a continuous curve, you will see vividly that all values between 2 and 10 3, the function f(t) can be taken!

Don't be wordy, hehe.

Solution: From the fundamental inequality: f(x)=f(x)+1 f(x) 2, and the equal sign is obtained if and only if f(x)=1.

Because f(x)=1 belongs to [,3], so f(x)min=2 when f(x)=, f(x)=

When f(x)=3, f(x)=3+1 3=10 3 5 2, so f(x)max=10 3

So the range of f(x)=f(x)+1 f(x) is [2,10 3].

Namely. When x>3.

f(x)=x+1 x.

Obtained by the nike function.

f(x)min=f(1)=2

Again. f（>f(3)=3+1/3

When the value range is 0, fmin = root number a

When fx is equal to 1 fx, the minimum value is obtained, and the solution is fx=1 or -1 (rounded), then the minimum value of fx is 1+1=2, when fx is or 3, the value of one of fx is the maximum, and the maximum value of fx is 10 3 when fx=3

So the fx range is 2 to 10 3

From the title: f(x) is the odd function with the domain r and the period is 2f(-5 2)=(5 2+2).

f(-1/2)

f(1/2)

When x (0,1), fx = 2x(1-x).

Macro-f(1 2)=-2*1 2*(1-1 2)finch f(-5 2)=-1 2

I wish you progress in your studies and have fun every day!

Analysis: The first one should be [-1,0].

The key to finding the maximum value of a quadratic function is to determine whether it is increasing or decreasing within this interval.

Because y=x2-2x+2

So: x=-b 2a=1, so in (- 1) monotonically decreasing, in (1,+ monotonically increasing.

So the function is monotonically decreasing at [-1,0].

ymax=f(-1)=5,ymin=f(0)=2, so the range at [-1,0] is:[2,5].

Therefore, the only judgment function is [0,3) which decreases first and then increases, and the state size 3 is farther away from the symmetry axis.

So: ymin=f(1)=1;f(3)=5 so the range of the function in [0,3) is:[1,5).

f(x)=(4/5)^x-1

4/5)^x>0

The mould is clustered to. 4/5)^x-1>-1

Therefore, the value of the sail field is (-1.).is infinite).

When a,b>=0, a+b>=2*root (a+b). If and only if a=b the minimum value of 2 * root number (a+b) is obtained

This problem is based on the above theorem. fx=fx+(1 fx)>=2 If and only if fx=1 fx, i.e., fx=1, fx obtains a minimum value of 2 When fx=3, fx goes to the maximum value of 3 and 1 3

So the range of fx is [2,3 and 1 3].

Let f(x) be expressed on [0,3] as f(x)=ax+b, and the odd function as f(0)=0;aDon't ask for it yet.

Let f(x) be expressed on [3,6] as f(x)=cx 2+dx+e, because f(5)=3,f(5).'=0 [f(5) is the maximum, the derivative of this point is 0], f(6)=2, and the simultaneous equation is solved to give c=-1, d=10, e=-22. So let f(x) be expressed on [3,6] as f(x)=-x 2+x-22

From this equation f(3)=-3 2+30-22=-1

Since the expression f(x)=ax on [0,3] also conforms to the expression f(x)=ax at x=3, f(3)=a3=-1, a=-1 3.

So f(x) is expressed on [0,3] as f(x)=-x 3.

Then according to the fact that f(x) is an odd function, we can write:

f(x) is expressed on [-3,3] as f(x)=-x 3;

f(x) is expressed as f(x)=-x 2+x-22 on [3,6];

f(x) on [-6,3] the expression is f(x)=-[-x 2+x-22]=x 2-x+22 This is written in terms of f(-x)=-f(x).

Let : u=fx, then the range of fx is the domain of fu, so the range of fx is: [4 3,3].

∵f(x)=x²-4x

x²-4x+4-4

x-2)²-4

The image opening of the function f(x) is upward, and the axis of symmetry is x=2, i.e., when x [2,5), the function f(x) is monotonically decreasing, when x [1,2], the function f(x) is monotonically decreasing, and when x=2 there is a minimum value f(2)=-4

When x = 5 there is a maximum value f(5) = (5-2) 2-4 = 5 Since x [2,5) and 5 is the open interval, the value range of this function is [-4,5).

Analysis: The first one should be [-1,0].

The key to finding the maximum value of the quadratic sail mountain vertical function is to determine whether the state is increasing or decreasing in this interval.

Solution: Because y=x 2-2x+2

So: x=-b 2a=1, so at (- 1) monotonically decreasing, at (1, + monotonically decreasing.

So the function is monotonically decreasing at [-1,0].

ymax=f(-1)=5,ymin=f(0)=2, so the range at [-1,0] is:[2,5].

So the function decreases first and then increases at [0,3) and 3 is farther away from the axis of symmetry.

So: ymin=f(1)=1;f(3)=5 so the range of the function in [0,3) is:[1,5).