How to judge the monotonicity of the first derivative in the range between a certain number and posi

Your derivative has been calculated, just determine whether it is greater than zero or less than zero in the interval.

The denominator of 2a 2x (x 2+a 2) 2 is obviously greater than 0, regardless of the interval in which x is taken; The numerator is 2(a2)x, right? If so, then the derivative is greater than 0 when x>0, otherwise it is less than or equal to 0, for the interval you give (root number 3 3*a, positive infinity), obviously when a > 0, x takes the value in the interval, there is always x>0, hence y'>0, the function increases monotonically. The monotonicity in other intervals can be similarly obtained.

y'=2a^2x/(x^2+a^2)^2

If the first derivative is greater than zero, the function increases, if the first derivative is less than zero, the function decreases, and if the first derivative is equal to zero, the function takes the extreme value.

Let x1,x2 be in the interval (root[3] 3 a,+ and the x10 function is incremented, w 2 a 2 x2 (x2 2 + a 2) 2 - 2 a 2 x1 (x1 2 + a 2) 2

2 a^2 (a^4 x1 - a^4 x2 - 2 a^2 x1^2 x2 - x1^4 x2 + 2 a^2 x1 x2^2 + x1 x2^4))/((a^2 + x1^2)^2 (a^2 + x2^2)^2)

2 a^2 (a^4 ( x1 - x2) +2 a^2 x1 x2 (x2 - x1) +x1 x2 (x2^3 - x1^3)))/((a^2 + x1^2)^2 (a^2 + x2^2)^2)

2 a^2 (a^4 (x1 - x2) +2 a^2 x1 x2 (x2 - x1) +x1 x2 (x2 - x1) (x1^2 + x1 x2 + x2^2)))/((a^2 + x1^2)^2 (a^2 + x2^2)^2)

2 a^2 (x2 - x1) (a^4 - 2 a^2 x1 x2 - x1^3 x2 - x1^2 x2^2 - x1 x2^3))/((a^2 + x1^2)^2 (a^2 + x2^2)^2)

The denominator > 0, 2 a 2 (x2 - x1) >0, just judge (a 4 - 2 a 2 x1 x2 x2 - x1 3 x2 - x1 2 x2 2 - x1 x2 3), because (root number [3] a) 3 < x1 < x2,a 4 - 2 a 2 x1 x2 x2 - x1 3 x2 - x1 2 x2 2 - x1 x2 3).

a^4 - 2 a^2 x1 x2 + x1^3 x2 + x1^2 x2^2 + x1 x2^3)

a 4 - 2 a 2 a 2 3 + a 4 9 + a 4 9 + a 4 9 + a 4 9), replace x1, x2 with (root[3] a) 3, so when (root[3] a) 3 < x1 < x2, w > 0, the function increases monotonically.

The way you asked the question was wrong, give it to me first, and then you asked casually.

y'=2a 2x (x 2+a 2) 2 =0 Since the denominator (x 2+a 2) 2 is not 0, only the numerator is 0, 2a 2x=0 so x=0, and then according to a>0, a<0 to discuss the increase and decrease!

It would be better to convert the original function to 1-a2 (x 2+a2) first, avoiding the need to find a derivative of the numerator.

The bumps and inflection points are all based on the second-order guide, and it has nothing to do with the first-order guide.

The increase or decrease of monotonicity is a sufficient and necessary relationship with the positive and negative of the first derivative.

There is no sufficient, insufficient, necessary, or unnecessary relationship between a point whose first derivative is equal to 0 and the point being an extremum.

A point where the first derivative is equal to 0 may or may not be an extremum, and an extreme point may be a point where the first derivative is equal to 0 or a discontinuity, and it is clear that the discontinuity point does not necessarily have a derivative, so how can you say that the derivative is equal to 0、、、 so the above two have nothing to do with each other.

However, the second derivative can be used to determine whether a point where the first derivative is equal to 0 is an extreme point 、、、

If the first derivative is equal to 0 and the second derivative is not equal to 0, then it can be said that the store must be an extreme point, which can be strictly proved with the limit of number preservation 、、、

Correspondingly, if the first derivative is equal to 0 and the even derivative is not equal to 0, then it can be said that the store must be an extreme point; If the derivative of the even order is greater than 0, then the point is the minimum point, if it is negative, the maximum point, and the proof of the holding of the limit can also be used.

The derivative of a section is greater than zero, the constant increase is less than zero, the second derivative is greater than zero, the concave function is less than the zero convex function.

Summary. Monotonicity Why is the derivative greater than 0 function monotonically increasing.

can help you understand.

Is it possible to say that the original function is greater than zero or less than zero as long as it is a single increase or a single decrease.

No. The following situation can be directly explained, and there is no need to look at the others. Right. Good.

For any point x on the field of the derivative function bai, according to the definition of the du derivative, f'(x)=lim(h 0)[f(x+h)-f(x)] h>0When zhih>0, there is x+h>x

Then according to the limit of the number of the number of DAO

, there is [f(x+h)-f(x)] h>0 in a neighborhood of x, so f(x+h)-f(x)>0, i.e., f(x+h)>f(x).

Let x+h=x1,x=x2, then when x1>x2, f(x1)>f(x2), f(x) increases monotonically.

The same is true at h<0.

Solution: Your thinking is not wrong, just keep asking!

f'(x)=x²+ax+1

1) When a=0;

f'(x)=x²+1>0

Thus, the original function is monotonically increasing on r;

2) When a ≠0 and a -4<0, i.e., a (-2,0)u(0,2), f'(x)=(x+1 2a) +1-1 4a 1 Therefore, the original function increases monotonically on r;

3) When a≠0, and |a|2 o'clock, order: f'(x)=0, then:

x1,2=[-a (a -4)] 2, then:

x∈(-a-√(a²-4)]/2]u[[-a+√(a²-4)]/2，+∞f(x)↑

x∈(-a-√(a²-4)]/2，-a+√(a²-4)]/2)，f(x)↓

First, see whether the derivative function is continuous, within the definition range of the original function, if the derivative function is not continuous and the original function is continuous, then the discontinuity point of the derivative function may be the extreme point, of course, it is only possible.

Second, if the original function is continuous, the derivative is also continuous, and the equation with the derivative equal to zero has no solution, then the sign of the derivative is always the same. This shows that the original function is monotonic throughout the defined domain.

The domain of f(x)=lnx+x is (0,+ f'(x)=(lnx) in this domain'+（x）'=（1/x）+1.

The derivative function is continuous under the defined domain (0,+) of f(x). and under this definition field, f'(x)=(lnx).'+（x）'=(1 x)+1>0, then f(x) is monotonically incremented under the definition domain (0, +.

In fact, you don't have to bother to calculate this at all, you can know it at a glance. lnx is monotonically increasing at (0,+, and x is monotonically increasing in the range of real numbers. The sum of the two, of course, is monotonically increasing within the scope of the defined domain.

Follow-up: The original function is f(x)=lnx+x

Follow-up: I've already figured it out above. The domain of f(x)=lnx+x is (0,+

Under this definition field, f'(x)=(lnx).'+（x）'=（1/x）+1。The derivative function is continuous under the defined domain (0,+) of f(x). and under this definition field, f'(x)=(lnx).'+（x）'=(1 x)+1>0, then f(x) is monotonically incremented under the definition domain (0, +.

In fact, you don't have to bother to calculate this at all, you can know it at a glance. lnx is monotonically increasing at (0,+, and x is monotonically increasing in the range of real numbers. The sum of the two, of course, is monotonically increasing within the scope of the defined domain.

If the function has a monotonically decreasing interval, x>0, and f'(x)=1/x+ax-2

It can be found directly by the derivative of composite functions and exponential functions, or you can find the logarithm first and then the derivative.

Monotonicity depends on whether the derivative is greater than 0 or less than 0

If you ask directly, you can.

If the derivative is greater than 0, it will increase, if the derivative is less than 0, it will decrease, and it is strictly monotonic, if the derivative is 0, it is a stable point, and it may be an extreme point.

There are formulas for the derivation of exponential functions and power functions, and they are carried out according to the formulas.

1、y=(1/3)^x，y ' = (1/3)^x * ln(1/3)。

2、y=x^-1，y ' = - x^-2 。

As for monotonicity, aren't exponential functions and power functions ready-made?

1. Since 0<1 3<1, the function is monotonically decreasing in (0,+.

2. Since -1<0, the function decreases monotonically on (0,+.

y = 3^(-x), y'= -3 (-x)ln3 < 0, the function decrements at (0, positive infinity);

y = x^(-1), y'= - x (-2) = -1 x 2 < 0, the function decrements at (0, positive infinity).

f(x) in (a,b) monotonous:

If the function increases monotonically, then f ( 0 always holds, and the function increases monotonically, then f ( 0 always holds, then f (x) ≠ 0, then f (x) has no zero point in (a, b).

The derivative is the tangent slope of the curve, when the curve is monotonous, the tangent slope is always positive (or negative), in fact, the derivative is always positive (or negative), of course, the derivative cannot be 0.