Let the function f x x 3 1 2 x 2 , where the zero point is located in the interval 1,2 to find the a

Solution: From f(x)=x 3-(1 2) (x-2), so f(1)=1-(1 2) (1-2)=1-2=-1<0, f(2)=8-(1 2) (2-2)=8-1=7>0, from f(x)=x 3-(1 2) (x-2) is a continuous function, and y=x 3 is an increasing function, y=(1 2) (x-2) is a decreasing function, and y=-(1 2) (x-2) is an increasing function.

Therefore f(x)=x 3-(1 2) (x-2) is an increasing function.

Therefore the zero point of f(x) is the only one that exists.

Therefore, the interval where the zero point is located is (1,2).

f(x)=x 3-(1 2) (x-2) is a continuous function.

Because f(1)=-1<0; f(2)=8>0

x 3 is an increment function.

1 2) (x-2) is a subtraction function and -(1 2) (x-2) is an increasing function.

f(x)=x 3-(1 2) (x-2) is the additive function.

The zero point of f(x) is the only one that exists.

The zero point is in the range of (1,2).

Because f(1)<0; f(2)>0；If the function f(x) is a continuous function, then according to the principle of zero-point existence, there must be a zero point between the intervals (1,2).

x monotonically increasing (0,0)(1,1) when x=2(2,8).

1 2)*(x-2) monotonically decreasing greater than 0(2,1) when x=1(1,2).

So in (1, 2).

f(1)=-1<0

f(2)=8>0

f(x) is a continuous function, so there is x that belongs to (1,2) such that f(x)=0

f(x)=x 3+x 2+x-1 Because the y value of the zero point is 0, so let f(x)=0 see the nucleus of the dividing nucleus then x 3+x 2+x-1=0 then x 3+x 2+x=1 then x(x 2+x+1)=1 then x 2+x+1=1 x are the images on both sides of the equation, and the two images indicate that there is an intersection point p, and it is obvious that the function f(x)=x 3+x 2+x-1 is in the interval (0,1) and only the pure zero point is either absent or not.

Proof: This question should use the technique of combining numbers and shapes.

If this function has a zero point, then f(x)=3 x and f(x)=x 2 have only one intersection point on [-1,0].

f(x)=3 x is in the range of [-1,0] is [one-third,1], and the function increases monotonically; The range of f(x)=x 2 on [-1,0] is [0,1], and the function is monotonically decreasing.

So this function has only one zero point on the interval [-1,0] (don't believe you to draw).

The Fang Yin Crack method is as follows, please refer to it:

<> If it helps the stove to be closed, please take the cavity with it.

f(x) = log x + 2x-11, the domain of the function f(x) is (0, + lim f(x) = x 0, lim f(x) = x + f'(x)>0, so f(x) increases monotonically, so f(x) has and only one zero point at (0,+).

f(x) =0, that is, log x+2x = 11, when 01, log x increases slower than 2x, so if you want to be 11 on the right side of the equation, you should consider 2x=11, solve x=, substitute x=5 into the equation and find log x+2x = log 5+10 > 11, and then substitute 4, log x+2x = 2+8 = 10 < 11, so the preliminary determination of f(x) zero is (4,5).

If you want to make a further determination, take the mean of 4 and 5 and substitute it, log x+2x = 2log 3+8, and the valuation operation of log 3 is calculated as, and the value obtained by substitution is > 11, so the range of the zero point of f(x) is (4,.

For this problem, it is sufficient to get the zero range as (4,5).

Solution: Method 1:

Derive it. f'(x)=(2^x)ln2+2x??>0 so f(x) is.

Monotonic increment function.

Then there is f(x) and (1,1) in x (0,1), so there is only one zero point.

Method 2: Because f(x)=2 x and f(x)=x3 are monotonic increment functions.

Therefore, the sum of the two increasing functions is also an increasing function in the definition, i.e., f(x)=2 x+x 3-2 in the interval x (0,1) is also an increasing function.

Then there is f(x) and (1,1) in x (0,1), so there is only one zero point.

Above! Hope it helps!

f'(x)=2^x*ln2+3x^2

When 00 i.e. f(x) is monotonically increasing on (0,1).

and f(0)=1+0-2=-1<0

f(1)=2+1-2=1>0

Then there is a point x0 in the interval (0,1) so that f(x0)=0, that is, f(x)=2 x+x 3-2 in the interval (0,1) is 1

Since the function f(x)=2x+x3-2 increases monotonically in the interval (0,1), and f(0)=-1 0, f(2)=10 0, so f(0)f(2) 0, so the function f(x)=2x+x3-2 has a unique zero point in the interval (0,1).

First, the derivation of f(x) is obtained.

The f(x) derivative is equal to 3 xln3+3x 2

This function is evergreen to zero.

So the function f(x) is.

The increment function will be brought in f(x).

f（0）=-1

f（1）=2

f(0)*f(1)<0

So there is only one zero point.

Proof: This question should use the technique of combining numbers and shapes.

If this function has a zero point, then f(x)=3 x and f(x)=x 2 have only one intersection point on [-1,0].

f(x)=3 x is in the range of [-1,0] is [one-third,1], and the function increases monotonically; The range of f(x)=x 2 on [-1,0] is [0,1], and the function is monotonically decreasing.

So this function has only one zero point on the interval [-1,0] (don't believe you to draw).

Combination of numbers:

The zero point of f(x) is the abscissa of the intersection of the function g(x)=2 x and h(x)=4 3+x images;

As can be seen from the figure, it is clear that g(x) and h(x) have two intersections; and a(x1,y1),b(x2,y2); To determine the approximate interval where x1 and x2 are located, 1h(-1)), 1h(2))).

You can use a geometry sketchpad or MATLAB or other drawing software to create a picture.

There is a zero point, located in the first quadrant, and the approximate interval is (3,4).

Select bf(0)=1-0=1 0

f(1 3)=(1 3) (1 3)-(1 3) (1 roll-off 2) exponential function bridge stupid g(x) = (1 3) The property of x gives f(1 3) 0

f(1 2) = (1 big 3) (1 2)-(1 2) (1 2) function g(x) = the property of x gives g(1 3) g(1 2).

i.e. f(1 2) 0

f（1）=-2/3＜0

f（2）=1/9-√2＜0

g(x)=1-x+x²/2-x³/3+……x 2013 2013f(x)+3=0 or g(x)-3=0

h(x)=f(x)+3=4+x-x²/2+x³/3-……x^2013/2013

h'(x)=1-x+x^2-..x 2012x=-1, h'(1)=2013>0

X≠1, h'(x)=1-x+x^2-..x^2012=(-x)^2013-1]/[x)-1]=(x^2013+1)/(x+1)

x>-1,h'(x)=(x^2013+1)/(x+1)>0x<-1,h'(x)=(x^2013+1)/(x+1)>0∴h'(x)>0 is constant, and h(x) is the increasing function.

h(0)=4

h(-1)=3+1-1-1/2-1/3-1/4-..1/2013∴h(-1)<0

f(x)+3=0Only 1 real solution belongs to (-1,0)i(x)=g(x)-3

Equally ego i'(x)=-1+x-x^2+..x 2012 <0i(x) is a subtraction function.

i(0)=-2<0

i(-1)=-3+(1+1+1/2+1/3+..1 2013)>0 g(x)-3=0, only 1 solution belongs to (-1,0) f(x)=0, and the real number is in the interval (-1,0).

The minimum value of b-a is 1