Answer to Question #177966 in Calculus for Abhijeet Pundir

Question

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=1 

Answer 1

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