anonymous
  • anonymous
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?
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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: \[m=\frac{ f(2+h)-f(2) }{ h }=\frac{ 0 }{ h }=0\]
anonymous
  • anonymous
Which only happnes if the line is horizontal

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anonymous
  • anonymous
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|
anonymous
  • anonymous
Thank you guys! (:
anonymous
  • anonymous
The only way the change in y is zero is if y=constant at that instant :)
anonymous
  • anonymous
Thus, line is horizontal

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