Derivative: the derivative value of a point on the curve is the slope of the tangent line of the point on the curve. I don't understand this explanation. Please answer briefly

Derivative: the derivative value of a point on the curve is the slope of the tangent line of the point on the curve. I don't understand this explanation. Please answer briefly

The derivative value is the function value when △ x approaches 0. A straight line intersects the curve. When moving the straight line, the two intersections gradually approach and then approach until the two intersections become a point, the straight line is a tangent, and the slope of the original straight line becomes the slope of the tangent

Given a ∈ R, the function f (x) = a X + LNX − 1, G (x) = (lnx-1) ex + X (where e is the base of natural logarithm) (1) Find the minimum value of function f (x) on interval (0, e]; (2) Is there a real number x0 ∈ (0, e], so that the tangent of curve y = g (x) at point x = x0 is perpendicular to the Y axis? If yes, find the value of x0; if not, explain the reason

（1）∵f(x)＝a
x+lnx−1，
∴f′(x)＝−a
x2+1
x＝x−a
x2
Let f '(x) = 0 to get x = a
① If a ≤ 0, then f '(x) > 0, f (x) monotonically increases in the interval (0, e], and the function f (x) has no minimum value
② If 0 < a < e, when x ∈ (0, a), f '(x) < 0, the function f (x) decreases monotonically on the interval (0, a),
When x ∈ (a, e], f '(x) > 0, the function f (x) increases monotonically on the interval (a, e],
So when x = a, the function f (x) takes the minimum value LNA
③ If a ≥ e, then f '(x) ≤ 0, the function f (x) decreases monotonically on the interval (0, e],
So when x = e, the function f (x) takes the minimum value a
e．
To sum up, when a ≤ 0, the function f (x) has no minimum value in the interval (0, e];
When 0 < a < e, the minimum value of function f (x) on interval (0, e] is LNA;
When a ≥ e, the minimum value of function f (x) on interval (0, e] is a
e．
（2）∵g（x）=（lnx-1）ex+x，x∈（0，e]，
∴g'（x）=（lnx-1）′ex+（lnx-1）（ex）′+1=ex
x+(lnx−1)ex+1＝(1
x+lnx−1)ex+1．
As can be seen from (1), when a = 1, f (x) = 1
x+lnx−1．
At this time, the minimum value of F (x) in the interval (0, e] is ln1 = 0, i.e. 1
X + LNX − 1 ≥ 0. (10 points)
When x0 ∈ (0, e], ex0 > 0, 1
x0+lnx0−1≥0，
∴g′(x0)＝(1
x0+lnx0−1)ex0+1≥1＞0．
The tangent of curve y = g (x) at point x = x0 is perpendicular to the Y axis, which is equivalent to equation G '(x0) = 0 and has a real number solution. (13 points)
G '(x0) > 0, that is, the equation G' (x0) = 0 has no real solution. Therefore, there is no x0 ∈ (0, e], so that the tangent of curve y = g (x) at point x = x0 is perpendicular to the Y axis

Known function f (x) = ax ^ 2 + X / e-lnx (where a is a constant and E is the base of natural logarithm) If f (x) is a monotonically decreasing function in its definition domain, find the value range of real number a When a > 0, there is no real solution to the equation f (x) = 0

Using the derivative method
f(x)‘=2ax+1\e-1\x

Given the function f (x) = e ^ x-ax-1 (a > 0, e is the base of the natural logarithm), if FX is greater than or equal to 0, it is constant for any X. find the value of the real number a

F (x) ≥ 0 is constant, that is, e ^ x ≥ ax + 1 is constant, draw the images of y = e ^ X and y = ax + 1. E ^ x ≥ ax + 1 is constant, that is, the image of y = e ^ x is above the image of y = ax + 1, and the images of these two functions pass through the point (0,1). Therefore, to make the image of y = e ^ x above the image of y = ax + 1, the straight line y = ax can only be connected with the curve y = e

Given the function f (x) = e ^ x-ax-1 (a > 0.. e is the base of the natural logarithm), find the minimum value of the function FX. If FX is greater than or equal to 0, it is constant for any X. find the value of the real number a

(1) F '(x) = e ^ x-a, Let f' (x) = 0, get e ^ x = a, x = LNA
It is easy to know that when x0, the minimum value of F (x) is f (LNA) = a - alna-1
(2) F (x) ≥ 0 is constant, which is equivalent to the minimum value f (LNA) ≥ 0, that is, a-alna-1 ≥ 0
Let g (a) = a-alna-1, then G '(a) = 1-lna - 1 = - LNA,
When 0

The function f (x) = 2lnx-x2 is known, if the equation f (x) + M = 0 is in [1] e. If there are two unequal real roots in e], the value range of real number m is __ (E is the base of natural logarithm)

∵f′（x）=2(1−x)(1+x)
x，
When x ∈ [1
e. 1), f '(x) > 0, f (x) is [1]
e. 1) is an increasing function,
When x ∈ (1, e), f '(x) < 0, f (x) is a subtractive function in (1, e),
When x = 1, f (x) has a maximum value, which is also the maximum value, f (1) = - 1,
F (1)
e）=-2-1
e2，f（e）=2-e2，
∴-2-1
e2≤-m＜-1，
∴1＜m≤2+1
e2．
So the answer is: (1, 2 + 1)
e2]．