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Pirate Bill · 06-18-2008, 01:23 PM

Solving for y1=x*7 at x=3; y=21.
Solving for y2=10^(x-1) at x=3; y=100.

Slope:
Derivative of y1: y1'=7 at x=3 (and all values of x)
Derivative of y2: y2'=10^(x-1); at x=3, y2'=100
Second derivative of y2: y2''=10^(x-1)

y1<y2 at x=3
y1'<y2' at x=3
y2'' is positive for all values of x indicating that y2' increases as x increases
therefore y1<y2 for all integers where x>2

That's way more complicated than just a pretty picture showing the plot of each equation on one graph, but I think it satisfies the burden of proof.

06-18-2008, 01:23 PM	#15
Pirate Bill Quality since 1973 Join Date: Dec 2007 Location: Right here Posts: 473	Solving for y1=x*7 at x=3; y=21. Solving for y2=10^(x-1) at x=3; y=100. Slope: Derivative of y1: y1'=7 at x=3 (and all values of x) Derivative of y2: y2'=10^(x-1); at x=3, y2'=100 Second derivative of y2: y2''=10^(x-1) y1<y2 at x=3 y1'<y2' at x=3 y2'' is positive for all values of x indicating that y2' increases as x increases therefore y1<y2 for all integers where x>2 That's way more complicated than just a pretty picture showing the plot of each equation on one graph, but I think it satisfies the burden of proof.
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