Let the function f x 2 x x 4, then the equation f x 0 must exist and what is the interval of the fol

f(x)=0<=>2^x=4-x

Let f(x) = 2 x

g(x)=4-x

Drawing f(x) and g(x) in the xoy coordinate system shows that the intersection points are in the interval (0,4).

Of course, the range can be narrowed, depending on how the question is asked!

The above is the solution for elementary mathematics.

If you use advanced mathematics.

Specifically, the zero-point existence theorem of a continuous function over a closed interval is used.

The detailed solution is as follows:

Because. f(1)=-1<0

f(2)=2>0

and f(x) is continuous over the closed interval [1,2].

Therefore, from the zero-point existence theorem, there must be a point & in the open interval (1,2), so that f(&)=0;

That is, f(x) must have a root in the interval (1,2).

Addendum: If the question is to prove that there is only one root in the interval (1,2).

The monotonicity of a function should be proved by derivatives.

The proofs are as follows: f'(x)=2 >0 , 1 so f(x) has only one root at f(1,2).

This problem is relatively simple to solve using the image method.

To make a Y-X image, the solution is as follows:

First, draw an image with y=2 x in the coordinate system.

Secondly, draw an image of y=-x+4 in the coordinate system.

Then the abscissa of the intersection of the two graphs is the answer.

First seek to define the domain, which is really the scope.

In derivatives, elementary functions, and the range of real numbers are continuously derivable.

f(0)=1+0-4=-3<0;

f(2)=4+2-4=2>0

Then there must be at least one value between 0 and 2 such that the equation f(x) = 0.

Generally, the * sign indicates the multiplication sign, while the slash table divides by or fraction hall profiles.

When x=1, f(x)=6, and when x=2, f(x)=8

How can you pretend that it is reduced?

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

x^2-3x)+2

x 2-3x+9 with jujube dust Peizen 4)+2-9 4(x+3 rock belt 2) 2-1 4 >=1 4, so f(x)>=1 4

Because x is not laughing 0, f(x)=1 (x+4+1 x).

Because x> hits the basis such as 0So x+1 x+4>=2+4=6, and only if x=1 is true. So 1 (x+4+1 x).

Derivative of f(x) yields g(x)=-(4x 2-16x+7) (2-x) 2;

Let g(x)>0, i.e. (2x-7)(2x-1)<0 yield, and x in [0 1].

So f(x) increases on [0 and decreases on [1].

f(-1)= -5/2<0

f(0)=1>0

So the range of 0 is (-1,0).