Answer to Question #177966 in Calculus for Abhijeet Pundir

Question #177966

determine the points of discontinuity of the dunction and the nature of discontinuity at each of these points f(x)={-x^2 when x<=0; 4-5x when 0<x<=1; 3x-4x^2 when 1<x<2; -12x+2x when x>2, also check whether function is drivable at x=1Β 


1
Expert's answer
2021-04-13T13:35:12-0400

Let us determine the points of discontinuity of the function "f" and the nature of discontinuity at each of these points.

"f(x)=\\begin{cases}-x^2,\\ \\ x\\le 0; \\\\4-5x,\\ \\ 0<x\\le1; \\\\3x-4x^2,\\ \\ 1<x\\le2; \\\\-12x+2x,\\ \\ x>2. \\end{cases}"

The function "f" is continuous on the open intervals "(-\\infty, 0),\\ (0,1), \\ (1,2)" and "(2,+\\infty)" because it coinсides with polynomial functions on these intervals.


Since "\\lim\\limits_{x\\to 0-}f(x)=\\lim\\limits_{x\\to 0-}(-x^2)=0=f(0)" and "\\lim\\limits_{x\\to 0+}f(x)=\\lim\\limits_{x\\to 0+}(4-5x)=4\\ne \\lim\\limits_{x\\to 0-}f(x)", we conclude that "x=0" is a point of non-removable discontinuity.


Since "\\lim\\limits_{x\\to 1-}f(x)=\\lim\\limits_{x\\to 1-}(4-5x)=-1=f(1)" and "\\lim\\limits_{x\\to 1+}f(x)=\\lim\\limits_{x\\to 1+}(3x-4x^2)=-1= f(1)", we conclude that the function "f" is continuous at the point "x=1".


Since "\\lim\\limits_{x\\to 2-}f(x)=\\lim\\limits_{x\\to 2-}(3x-4x^2)=-10=f(2)" and "\\lim\\limits_{x\\to 2+}f(x)=\\lim\\limits_{x\\to 2+}(-12x+2x)=-20\\ne \\lim\\limits_{x\\to 2-}f(x)", we conclude that "x=2" is a point of non-removable discontinuity.


Let us check whether function is differentiable at "x=1."

Since "\\lim\\limits_{x\\to 1-}\\frac{f(x)-f(1)}{x-1}=\\lim\\limits_{x\\to 1-}\\frac{4-5x+1}{x-1}=\n\\lim\\limits_{x\\to 1-}\\frac{-5(x-1)}{x-1}=-5" and "\\lim\\limits_{x\\to 1+}\\frac{f(x)-f(1)}{x-1}=\\lim\\limits_{x\\to 1+}\\frac{3x-4x^2+1}{x-1}=\n\\lim\\limits_{x\\to 1+}\\frac{(x-1)(-1-4x)}{x-1}=\\lim\\limits_{x\\to 1+}(-1-4x)=-5", we conclude that the function "f" is differentiable at "x=1."


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