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monroe17

  • 2 years ago

True or False and explain why? For a differentiable function y=f(x), f'(2)=0 means that the tangent line to the graph of f at x=2 is horizontal. I know it's true, but how do I explain why?

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  1. Sujay
    • 2 years ago
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    A derivative gives the slope of a tangent point for any point where it applies. So since the slope at x=2 would be 0, that means it must be a horizontal line.

  2. quantum77
    • 2 years ago
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    Use the definition limit definition of a derivative: If thederivative is 0 then \[f \prime(2)=\lim_{h \rightarrow 0}=\frac{ f(2+h)-f(2) }{ h }\] If this is zero than This implies that f(2+h)-f(2)=0, so thinking about the defintion of the slope: \[m=\frac{ f(2+h)-f(2) }{ h }=\frac{ 0 }{ h }=0\]

  3. quantum77
    • 2 years ago
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    Which only happnes if the line is horizontal

  4. eseidl
    • 2 years ago
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    you could prove this using the definition of the derivative:\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\]Another way to say this is:\[f'(x)=\lim_{\Delta x \rightarrow 0}\frac{\Delta y}{\Delta x}=\frac{dy}{dx}\]Graphically, this is the slope of the tangent line at the point as @Sujay said.|dw:1354847377868:dw|

  5. monroe17
    • 2 years ago
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    Thank you guys! (:

  6. eseidl
    • 2 years ago
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    The only way the change in y is zero is if y=constant at that instant :)

  7. eseidl
    • 2 years ago
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    Thus, line is horizontal

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