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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)...
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.
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
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.