The number of zeros of the function f x x 2 x

x+2^x=0

That is: 2 x = -x

That is, the number of intersections between y1=2 x and y2=-x images.

Sketch the two functions, and it is easy to get that y1 and y2 have only one intersection point.

So, the number of zeros of f(x)=x+2 x is 1.

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

1 pc. Proof:

f(0)=1 when x=0 f(-1)=-1 when x=-1 2f'(x)=1+2 x *ln2>0 is constant.

So f(x) is monotonically increasing on r.

Because f(0)>0 f(-1)<0

So f(x) has a zero point on r, between (-1,0).

If you don't want to use rigorous proofs, you can draw a graph of the function directly, which is faster to choose a fill-in-the-blank question

If you don't understand, you can ask, hope

f'(x)=(x+2^x)'

x'+(2^x)'

1+(2^x)log2

From the derivative of this function, it can be seen that the secondary function is monotonically increasing.

x+2^x=0

x=-2^x

x=-wn(log2)/log2(n=)

The calculation shows that when x is approximately equal to , f(x)=0

So there is 1 zero point.

Anranlethe's solution is currently the best solution to this problem, and it's very good.

That is, there are several solutions to x2-2 x=0.

That is, find how many intersections y=x 2 and y=2 x have.

Draw an image of the two functions.

You can see x=2

x=4 is obviously.

And the two functions return to x" There is an intersection point at the senji at 0 of this liquid.

So the answer is 3.

It is the preliquid height f(x) = 2x-x 2

Is it? If so, you can do it with the following method:

Let f(x)=0, i.e. 2x-x 2=x(2-x)=0, so x=0 or x=2

So f(x)=2x-x 2

There are two of the zero points of the hail.

The number of zeros, which is the number of x's that make f(x) zero.

This equation is a transcendental equation that you can't solve if you're in high school. If you use the dichotomy, you can find the approximate root. To draw, you can use the Geometry Artboard.

One way to do this is to think of the zero point of the function as the intersection of f(x)=x 2 and f(x)=2 x (since x 2-2 x=0 can be transformed into x 2=2 x), and roughly draw a rough image of the two functions on the coordinates, and you can see that there are three intersections.

x 0 is obviously the zero point of x=.

Draw to know. x<0x and 2 x also have an intersection.

So there are 3 of them.

It's too cumbersome to draw as you can imagine.

f（x）=2^x-x^2=0

2^x=x^2

Let y1=2 x y2=x 2

It is the number of intersections of two functions.

Draw a simple diagram, easy to know = 2

First find the first derivative of f(x), then make the first derivative equal to 0, and solve the value of x!

Obviously x= is the zero point.

Draw to know. At x<0, x and 2 x have an intersection.

So there are 3 of them.

x^2-2^x=0

x^2=2^x

x=2, x=4 satisfies.

Paint x=-1 x 2=1, 2 x=1 2x=0 x 2=0, 2 x=1

So there are 3 intersections on [-1,0].

f(x)=(x^3-1)+(x-1)=(x-1)*(x^2+x+1)+(x-1)=(x-1)*(x^2+x+2)=0

Getting x=1 means that there is only one zero point that is x=1

Derivative f"(x)=3x2+1>0, so f(x) is a monotonic increasing function, f(-1)<0, f(2)>0, so there is and only one point x, such that f(x)=0