The number of real roots of the equation x ^ 2 = 2 ^ x is

Let Y1 = x ^ 2, y2 = 2 ^ X
Drawing the images of Y1 and Y2, it is easy to find that there is only one intersection point in the second quadrant,
The first quadrant is two intersections
In fact, the intersection of the first quadrant can be found
When x = 2, Y1 = 4, y2 = 4
When x = 4, Y1 = 16, y2 = 16
Therefore, the number of real roots of the original equation is three
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Quadratic function and derivative
There is an inscribed rectangle ABCD in the graph enclosed by the image of quadratic function f (x) = 4x-x ^ 2 and X axis. Find the maximum area of this rectangle

Let a point coordinate (x, 4x-x ^ 2), B point coordinate (x, 0) C (4-x, 0) d (4-x, 4x-x ^ 2), then the rectangular area s = 2 (2-x) * (4x-x ^ 2) x is greater than 0 and less than 2. Then, take the derivative of S, calculate x when s' = 0, round off the unsatisfied part of X, and bring back the formula of S

How to find the derivative of quadratic function

Let the quadratic function be y = ax ^ 2 + BX + C
Then y '= (AX ^ 2 + BX + C)'
=(ax^2)'+(bx)'+c‘
=2ax+b

The relationship between △ and monotone interval when derivative is quadratic function
What about △ when there are exactly three monotone intervals? Why can't two be equal?

When △≤ 0, the quadratic function is constant ≤ 0 (or ≥ 0), because it is a monotone interval
When △ > 0, the intersection of quadratic function and X axis, so there are three intervals with different signs, so the original function has three monotone intervals

How to judge the number of zeros by derivative? Or what is the general method to judge the number of zeros

Derivative can not determine the solution of the original function, that is, the zero point
To judge the zero point, you can take any value of the original function. If it is greater than zero and less than zero, there will be zero point
For example, the function y = x * x * x-3x-3
X = 2, y = - 10, then there must be a root in between. Of course, combined with the monotonicity of the function (or the positive and negative interval of the derivative), we can roughly judge the function image. However, only knowing the derivative, we can not determine the number of zero points, we still need to use the value and extreme value

The derivative of arctanx at zero

f'(x)=(arctanx)'=1/(1+x^2)
f'(0)=1

If the function f (x) = | x | + 1 + K has two zeros, then the value range of real number k is______ ．

From the meaning of the question, we can get that there are two intersection points of the image and the line y = - K of the function y = | x | + 1, and the combination of numbers and shapes can get - k > 1, K < 1, so the answer is (- ∞, - 1)

The number of zeros of the function f (x) = lnx-1 / (x-1) is
A. 3 B.2 C.1 d.0

Draw the images of y = LNX and y = 1 / (x-1) respectively. As shown in the figure, it is easy to see that the two functions have two intersections, that is, the number of zeros of function f (x) = lnx-1 / (x-1) is 2
Choose B

F (x) = 1 / X * SiNx ^ 2, find the derivative f (Π / 2)

f'(x)=（2xsinxcosx-sinx^2）╱x∧2...∴f'（π╱2）=-4╱π∧2

The number of real roots of the equation x ^ 2 = 2 ^ x is