## opendiscuss 3 years ago The slope of the graph of f(x) = |x| changes abruptly when x = 0. Does this function have a derivative? If so, what is it? If not, why not? Explanation is given in the course material, but not clear...Why cant the tangent line be defined at (0,0)...

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1. iisthphir

The function is not continuous at 0 so the derivative can't be taken. The two sided limit as x > 0 does not exist. As the function is elsewhere linear, its derivative is constant. For x < o it is -1 for x > 0 it is 1.

2. khoutir

the function is continuous but hasn't a derivative because: $\lim_{x \rightarrow -0}[f(x+\Delta x)-f(x)/\Delta x] \neq \lim_{x \rightarrow +0}[f(x+\Delta x)-f(x)/\Delta x]$ for x<0 we have :$f(x+\Delta x)-f(x)/\Delta x=[-(x+\Delta x)+x]/\Delta x= -1$ for x>0 we get :$f(x+\Delta x)-f(x)/\Delta x=[(x+\Delta x)-x]/\Delta x= 1$ so the limit of the difference quotient doesn't exist and so that the derivative at x=0

3. JingleBells

f(x)=|x| is continuous at x=0: $f(x ^{-})=\lim_{x \rightarrow 0}\left| x \right|=0$ $f(x ^{+})=\lim_{x \rightarrow 0}\left| x \right|=0$ |dw:1342611754796:dw| However, $f \prime(x)$ has a point of jump discontinuity at x=0: $f \prime(0^{-})=\lim_{x \rightarrow 0}-1=-1$ $f \prime(0^{+})=\lim_{x \rightarrow 0}1=1$ |dw:1342612014321:dw| at x=0 left hand limit $\neq$ right-hand limit as Problem Set 1 Question 1D, 3 (d) and (e) demonstrate similar cases.