What is the relationship between the image of a function and the image of its derivative?

What is the relationship between the image of a function and the image of its derivative?

The derivative is the slope of the function at that point. If the derivative is greater than 0, the function will increase monotonically and decrease monotonically. You should know this. The following is a simple example. I'll ask you where it is not clear~

If the function f (x) whose derivative function is y = x ^ (1 / 2) passes through the point (9,1), then the function f (x)=

If you are a senior high school or a university, you just need to take (9,1) into the integral. If you are a senior high school, if you are given a function, you can calculate its derivative. If you are given a function's derivative, can you calculate the function, because the constant term will be eliminated in the process of derivation. When you integrate, you must add an unknown constant after the function you calculated, and then (9,1) will be added, 1) If you replace it, you can get this constant. If you do it, you can do it yourself

What function has a derivative of 0?

A constant function, such as y = 34

If there is infinite derivation, the derivative at a certain point is always a function of 0, which is a very important function

Calculate all derivatives of F (x) = e ^ {- x ^ 2} - 1 at x = 0

The derivative function f '(x) = a (x + 1) (x-a) of function f (x) is known. If f (x) reaches the minimum at x = a, then the value range of real number a is______ ．

From F ′ (x) = a (x + 1) (x-a) = 0, the solution is a = 0 or x = - 1 or x = A. if a = 0, then f ′ (x) = 0. If a = - 1, then f ′ (x) = - (x + 1) 2 ≤ 0. If a ＜ - 1, then f ′ (x) = - (x + 1) 2 ≤ 0

A derivative problem of mathematical function in college entrance examination
Given the function f (x) = LNX, G (x) = k · (x -!) / (x + 1)
(II) when x > 1, f (x) > 1
G (x) is constant, and the value range of real number k is obtained;
(III) let positive real numbers A1, A2, A3 and an satisfy a1 + A2 + a3 +... + an = 1,
Verification: ln (1 + 1 / A1 & # 178;) + ln (1 + A2 & # 178;) +. + ln (1 + 1 / an & # 178;) > 2n & # 178; / (n + 2)
The answer to (II) is (- ∞, 2),
It is known from (2) that it holds when INX > 2 · (x-1) / (x + 1)
Let x = 1 + 1 / an & # 178; (0 < an < 1), then in (1 + 1 / an & # 178;) > 2 / (2An & # 178; + 1) > 2 / (2An + 1)
So ln (1 + 1 / A1 & # 178;) + ln (1 + 1 / A2 & # 178;) +. + ln (1 + 1 / an & # 178;) > 2 (1 / (2A1 + 1) + 1 / (2A2 + 1) +... 1 / (2An + 1))
And 2 (1 / (2A1 + 1) + 1 / (2A2 + 1) +... 1 / (2An + 1)) [(2A1 + 1) + (2A2 + 1) +... (2An + 1)] ≥ n & # (how to do this step, please explain in detail,
SO 2 (1 / (2A1 + 1) + 1 / (2A2 + 1) +... 1 / (2An + 1)) ≥ 2n & # 178; / (n + 2)
So: ln (1 + 1 / A1 & # 178;) + ln (1 + A2 & # 178;) +. + ln (1 + 1 / an & # 178;) > 2n & # 178; / (n + 2)

It is known that FX is a differentiable function on R
If we know that FX is a differentiable function on R, and for any x belonging to R, f (x) > F '(x), then
A.e^2013·f(-2013)e^2013·f(0)
B.e^2013·f(-2013)e^2013·f(0)
D.e^2013·f(-2013)>f(0),f(2013)

The answer is d
Just find a special function to replace it
For example, G (x) = e ^ X
g'(x) = g(x)
Let f (x) = g (x) + 1 = e ^ x + 1
Let f (x) > F '(x)
f(0) = 2
f(2013) = e^2013+1
e^2013 * f(-2013) = e^2013 * (e^-2013 + 1) = 1+ e^2013
obviously
e^2013 * f(-2013) > f(0)
f(2013) < e^2013 * f(0)
So the answer is d

Let f (x) = P (x-1x) - 2lnx, G (x) = X2, (I) if the line L is tangent to the image of function f (x), G (x) and tangent to the image of function f (x) at point (1, 0), find the value of real number p; (II) if f (x) is a monotone function in its domain, find the value range of real number P

(I) method 1: ∵ f '(x) = P + PX2 − 2x, ∵ f' (1) = 2p-2. Let a straight line and G (x) = x2 be tangent to the point m (x0, Y0) ∵ G '(x) = 2x, ∵ 2x0 = 2p-2, then the solution is ∵ x0 = P − 1, Y0 = (P − 1) 2, and the solution is p = 1 or P = 3 by substituting the straight line equation l into y = X2 to get 2 (p-1) (x-1) = 0, ∵ △ 4 (p-1) 2-8 (p-1) = 0, then the solution is p = 1 or P = 1= 3. (II) ∵ f '(x) = P + PX2 − 2x = PX2 − 2x + PX2.. ① if f (x) is a monotone increasing function, f' (x) ≥ 0 is constant at (0, + ∞), that is, px2-2x + P ≥ 0 is constant at (0, + ∞), that is, P ≥ 2xx2 + 1 = 2x + 1x is constant at (0, + ∞), and 2x + 1x ≤ 1, then f (x) is a monotone increasing function at (0, + ∞) when p ≥ 1; ② In order to make f (x) a monotone decreasing function, f '(x) < 0 must be constant at (0, + ∞), that is, P ≤ 2xx2 + 1, (0, + ∞), and 2xx2 + 1 ≥ 0, so p ≤ 0. When p ≤ 0, f (x) is a monotone decreasing function at (0, + ∞). In conclusion, if f (x) is a monotone function at (0, + ∞), then the value range of P is p ≥ 1 or P ≤ 0

The function f (x) = x2 + 2x + alnx (a ∈ R) is known. When t ≥ 1, the inequality f (2t-1) ≥ 2F (T) - 3 holds, and the value range of real number a ()
A. （-∞，1）B. （-∞，2）C. （-∞，1]D. （-∞，2]

∵ f (x) = x2 + 2x + alnx (a ∈ R). When t ≥ 1, the inequality f (2t-1) ≥ 2F (T) - 3 is tenable, ∵ 2t2-4t + 2 ≥ alnt2 AlN (2t-1) ∵ 2t2-alnt2 ≥ 2 (2t-1) - AlN (2t-1) let H (x) = 2x alnx (x ≥ 1), then the problem can be reduced to h (T2) ≥ H (2t-1) ∵ t ≥ 1

What is the meaning of the derivative of the derivative of the cubic equation of one variable
If the two sides of x = 1 are different, why should the derivative x = 1 of the equation be equal to 0
Can you answer o (∩) in detail_ Thank you
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The derivative of the equation of one variable is the corresponding slope, right
So the derivative of his derivative is the rate of change of the slope
If the slope of a function is increasing all the time
So the derivative of his derivative is a positive value
If the slope of a function is a constant value,
So the derivative of his derivative is zero, because his slope doesn't change