運算式 7: "m" Subscript, 2 , Baseline equals StartFraction, 2.9 4 8 minus 2.8 1 4 Over 4.2 minus 4.0 , EndFractionm2=2.948−2.8144.2−4.0
equals=
0.6 70.67
7
運算式 8: "l" Subscript, 2 , Baseline left parenthesis, "x" , right parenthesis equals 2.8 1 4 plus "m" Subscript, 2 , Baseline left parenthesis, "x" minus 4.0 , right parenthesis left brace, 4.0 less than "x" less than 4.2 , right bracel2x=2.814+m2x−4.04.0<x<4.2
8
what is the right value of instantaneous rate of change, m? Well, it's smaller than m1 (represented in red) and larger than m2 (represented in orange).... adjust the slider below to see what I mean.
9
運算式 10: "m" equals 0.7 2m=0.72
00
22
10
運算式 11: "l" left parenthesis, "x" , right parenthesis equals 2.8 1 4 plus "m" left parenthesis, "x" minus 4.0 , right parenthesis left brace, 3.8 less than "x" less than 4.2 , right bracelx=2.814+mx−4.03.8<x<4.2
11
There's no right answer, but one good guess is: it's roughly m3 (represented in green)
12
運算式 13: "m" Subscript, 3 , Baseline equals StartFraction, 2.9 4 8 minus 2.6 6 1 Over 4.2 minus 3.8 , EndFractionm3=2.948−2.6614.2−3.8
equals=
0.7 1 7 50.7175
13
運算式 14: "l" Subscript, 3 , Baseline left parenthesis, "x" , right parenthesis equals 2.6 6 1 plus "m" Subscript, 3 , Baseline left parenthesis, "x" minus 3.8 , right parenthesis left brace, 3.8 less than "x" less than 4.2 , right bracel3x=2.661+m3x−3.83.8<x<4.2
