How can you tell if a function has several zeros? And how do you tell if a function has a zero point

The most intuitive way to determine the zero point of a function is to draw a graph.

Example: |x|=1+ax has a negative root and no positive root, find the value range of a.

x|=1+ax

Equivalent. x^2=(1+ax)^2

(a 2-1) x 2+2ax+1=0 has a negative root and no positive root, and then a 2-1 is discussed.

When a 2-1 = 0

That is, when a=1 and -1, they can be obtained by substituting the original formula respectively.

a=1 holds.

a=-1 is not true.

When a 2-1<0, this quadratic function is imaged due to (a 2-1) x 2+2ax+1.

(0,1), if the opening is downward, then the function must have an intersection point with the x positive half axis (the positive root appears, which contradicts the problem), so it is not true.

When a 2-1>0.

Combine images. delta>=0

b/2a<0

After the column is connected, a>1 can be solved

Then the 3 cases are combined.

a>=1

f(a)f(b)<=0 may appear in such problems, such as where a function has a root in x [a,b].

What he means is that the image has an intersection point at x [a,b]. Regardless of the direction of the opening, f(a) and f(b) must be one plus and one negative or one is zero and one is not a bell, so f(a) f(b) 0

I don't know if you see it?

If you understand, add some points.

The first step is to find the derivative of the function and judge its monotonicity. The second step is to determine whether the function has a zero point according to the monotonic interval. Of course, you don't give the specific function, you can only provide the solution idea, hope!

Here's how to determine the approximate interval where the zero point of the function is located:

Method 1: If the image of the function y=f(x) on the closed interval [a,b] is a continuous curve, and the sign of the function value at the end of the interval is different, i.e., f(a)·f(b) 0, then in the interval [a,b], the function y=f(x) has at least one zero point in the number of the universe, that is, the corresponding equation f(x)=0 has at least one real solution in the interval [a,b].

Method 2, the zero point of the function y=f(x) is the real root of the equation f(x)=0, that is, the abscissa of the intersection of the image of the function y=f(x) and the x-axis (straight line y=0), so the equation f(x)=0 has the real root, the image of the push-out function y=f(x) has an intersection point with the x-axis, and the push-out function y=f(x) has a zero point.

Method 3: The zero point of the function f(x)=f(x)-g(x) is the real root of the equation f(x)=g(x), that is, the abscissa of the intersection of the image of the function y=f(x) and the image of the function y=g(x), which is very useful.

Here's how to determine the approximate interval where the zero point of the function is located:

Method 1: If the image of the function y=f(x) on the closed interval [a,b] is a continuous curve, and the sign of the function value at the end of the interval is different, taking the cluster that is, f(a)·f(b) 0, then in the interval [a,b], the function y=f(x) has at least one zero point, that is, the corresponding equation f(x)=0 has at least one real solution in the interval [a,b].

Method 2, the zero point of the function y=f(x) is the real root of the equation f(x)=0, that is, the abscissa of the intersection of the image of the function y=f(x) and the x-axis (straight line y=0), so the equation f(x)=0 has the root of the real number, the image of the basis function y=f(x) has an intersection point with the x-axis, and the push-out function y=f(x) has a zero point.

Method 3: The zero point of the function f(x)=f(x)-g(x) is the real root of the equation f(x)=g(x), that is, the abscissa of the intersection of the image of the function y=f(x) and the image of the function y=g(x).

Method 1: Definition.

Steps: The first step is to judge the monotonicity of the function;

The second step is to verify whether the product of the pure branch imaginary value of the function at the end of the interval is less than 0 according to the existence theorem of the zero point. If its product is less than 0, then the interval is the existence of a unique zero point interval or the zero point is calculated directly by using the idea of the equation;

Step 3 Draw conclusions.

Example].The number of zeros of the function is ( ).

a．0 b．1 c．2 d．3

Analysis] is known.

Therefore, in is monotonically increasing, and , so the number of zeros of is 1, so choose b

Fang do burning method 2: number combination method.

Problem solving steps: The first step is to convert a zero-point problem into an equation with a root;

Step 2 In the same Cartesian coordinate system.

, draw the images of the functions and , respectively;

Step 3 Observe and judge the number of intersections of the image of the function and .

Step 4: The number of intersections of the and images is equal to the zero point of the function.

Example].The number of solutions to the equation is ( ).

a．3 b．2 c．1 d．0

Analysis] From the image, it can be seen that the function and the function have 2 intersections, so the equation has 2 solutions, choose b

y=xln(x+1) defines the domain x>-1

y'=ln(x+1)+x/(x+1) (uv)'=u'v+uv'

y''=1 (x+1)+[x+1-x] (x+1) > Yun Min Spike 0 y'Monotonically increasing take late y'Up to one zero point.

y'(0)=0

y has and only one extreme point, and the extreme point x=0 is the minimum point (y''>0）y(0)=0

The function has only 1 zero point.

Here's how to determine the approximate interval where the zero point of the function is located:

Method 1: If the image of the function y=f(x) on the closed interval [a,b] is a continuous curve, and the sign of the function value at the end of the interval is different, i.e., f(a)·f(b) 0, then in the interval [a,b], the function y=f(x) has at least one zero point in the number of the universe, that is, the corresponding equation f(x)=0 has at least one real solution in the interval [a,b].

Method 2, the zero point of the function y=f(x) is the real root of the equation f(x)=0, that is, the abscissa of the intersection of the image of the function y=f(x) and the x-axis (straight line y=0), so the equation f(x)=0 has the real root, the image of the push-out function y=f(x) has an intersection point with the x-axis, and the push-out function y=f(x) has a zero point.

Method 3: The zero point of the function f(x)=f(x)-g(x) is the real root of the equation f(x)=g(x), that is, the abscissa of the intersection of the image of the function y=f(x) and the image of the function y=g(x), which is very useful.

y=xln(x+1) defines the domain x>-1

y'=ln(x+1)+x/(x+1) (uv)'=u'v+uv'

y''=1 (x+1)+[x+1-x] (x+1) > Yun Min Spike 0 y'Monotonically increasing take late y'Up to one zero point.

y'(0)=0

y has and only one extreme point, and the extreme point x=0 is the minimum point (y''>0）y(0)=0

The function has only 1 zero point.

There is only one, x<0 to draw y=x 3.

y=-ln|x|x<0, the equation becomes the source number y=-ln-x plotted to see that the two vertical rough plots have only one intersecting crack point, so there is only one zero point.