## monroe17 Group Title 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? one year ago one year ago

1. Sujay Group Title

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 Group Title

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 Group Title

Which only happnes if the line is horizontal

4. eseidl Group Title

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 Group Title

Thank you guys! (:

6. eseidl Group Title

The only way the change in y is zero is if y=constant at that instant :)

7. eseidl Group Title

Thus, line is horizontal