Asked by Harminder
Determine the instantaneous rate of change at x=2 for the function defined by m(x) = -x^2 +2x
Answers
Answered by
Anonymous
If you know calculus
dm/dx = -2 x + 2 = -4 +2 = -2
dm/dx = -2 x + 2 = -4 +2 = -2
Answered by
PsyDAG
a minus times a minus = a positive
-x^2 +2x = 4 + 4 = 8
-x^2 +2x = 4 + 4 = 8
Answered by
oobleck
??
-x^2 ≠ (-x)^2
-x^2 ≠ (-x)^2
Answered by
Anonymous
If with algebra
m(x+dx) = -(x+dx)^2 + 2 x + 2 dx
= -x^2 -2 x dx -dx^2 + 2 x + 2 dx
m(x) = -x^2 +2x
subtract
change in m = -2 x dx -dx^2 + 2 dx
slope = ( -2 x dx -dx^2 + 2 dx)/dx
= - 2 x - dx + 2
for instantaneous slope find limit as dx --->0
-2x + 2
when x = 2
-4 + 2 = -2 (but of course we knew that)
m(x+dx) = -(x+dx)^2 + 2 x + 2 dx
= -x^2 -2 x dx -dx^2 + 2 x + 2 dx
m(x) = -x^2 +2x
subtract
change in m = -2 x dx -dx^2 + 2 dx
slope = ( -2 x dx -dx^2 + 2 dx)/dx
= - 2 x - dx + 2
for instantaneous slope find limit as dx --->0
-2x + 2
when x = 2
-4 + 2 = -2 (but of course we knew that)
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