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If f'(x) = ln(x-2), then the graph of y = f(x) is decreasing if and only if
a) 2< x < 3
b) 0
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If the first derivative of a function is equal to 0, what can you discern from that?
x = 2 ?
Let's try this differently, ok, so do you know what the graph of \[ ln(x)\] looks like?
yes |dw:1452372946319:dw|
Good! Okay so ln(x-2) is simply shifted 2 units to the right. And Trojan, that is incorrect.
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|dw:1452373034309:dw|
Therefore, do you agree that the asymptote of f'(x) is at x=2 and has a zero at x=3?
i know that the slope is positive from that function
f'(x)=0 at x=3 so that is where f(x) will change.
Trojan, substitute 2.5, would you?
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|dw:1452373306066:dw|
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