微分均值定理
Author: 藍邦偉  老師

$ O $

$ 微分均值定理 $

$ 若f(x)在(a,b)可微且在[a,b]連$
$續，\;則在(a,b)中至少存在一$
$數c\;，\;使得f'(c)=\frac{f(b)-f(a)}{b-a}. $

$ A $

$ B $

$ C $

u = Vector((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
O = (0, 0)
Q_2: 0.7 * Corner[1] + 0.3 * Corner[2]
Q_3: (x(Q_2), y(Corner[4]))
poly1, q_1, q_2, q_4, q_3 = Polygon(Corner[1], Q_2, Q_3, Corner[4])
n = 0 (slider from 0 to 5, step 1)
A_1 = (-2, 1)
A_2 = (1, 4)
A_3 = (4, 2)
A_4 = (7, 5)
f = Polynomial({A_1, A_2, A_3, A_4})
A = Point(f)
B = Point(f)
C = Point(f)
a: x(A)
b: y(A)
c: x(B)
d: y(B)
L = Tangent(C, f)
text1: "$ \frac{f(b)-f(a)}{b-a} =\frac{" + (LaTeX[d]) + "-" + (LaTeX[If[b < 0, "("]]) + "" + (LaTeX[b]) + "" + (LaTeX[If[b < 0, ")"]]) + "}{" + (LaTeX[c]) + "-" + (LaTeX[If[a < 0, "("]]) + "" + (LaTeX[a]) + "" + (LaTeX[If[a < 0, ")"]]) + "}$
$=\frac{" + (LaTeX[d - b]) + "}{" + (LaTeX[c - a]) + "}=" + (LaTeX[(d - b) / (c - a)]) + "$"
text2: "$ f'(c)=f'(" + (LaTeX[x(C)]) + ")=" + (LaTeX[f'(x(C))]) + " $"
e = Segment(A, B)