anonymous
  • anonymous
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)...
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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.

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