Answer to Question #185363 in Math for akshat

Question

determine the point of discontinuity of the function f and the nature of discontinuity at each 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

and check whether function f is derivable at x=1

Answer 1

$$f(x) = \begin{cases} -x^2 &\text{when } x\leq0 \\ 4-5x &\text{when } 0<x\leq1\\ 3x-4x^2 &\text{when } 1<x\leq2\\ -12x+2x &\text{when } x>2\\ \end{cases}$$

$$\lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^-}(-x^2)=0$$

$$\lim\limits_{x\to0^+}f(x)=\lim\limits_{x\to0^+}(4-5x)=4$$

$$\lim\limits_{x\to0^-}f(x)=0\not=4=\lim\limits_{x\to0^+}f(x)$$

$\lim\limits_{x\to0}f(x)$ does not exist.

The function $f(x)$ is discontinuous at $x=0$ and has a jump discontinuity at $x=0.$

$$\lim\limits_{x\to1^-}f(x)=\lim\limits_{x\to1^-}(4-5x)=-1$$

$$\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^+}(3x-4x^2)=-1$$

$$\lim\limits_{x\to1^-}f(x)=-1=\lim\limits_{x\to1^+}f(x)$$

$\lim\limits_{x\to1}f(x)=-1$.

$$f(1)=4-5(1)=-1$$

The function $f(x)$ is continuous at $x=1.$

$$\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(3x-4x^2)=-10$$

$$\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(-12x+2x)=-20$$

$$\lim\limits_{x\to2^-}f(x)=-10\not=-20=\lim\limits_{x\to2^+}f(x)$$

$\lim\limits_{x\to2}f(x)$ does not exist.

The function $f(x)$ is discontinuous at $x=2$ and has a jump discontinuity at $x=2.$

$$f(1)=-1$$

$$\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=\lim\limits_{h\to0^-}\dfrac{4-5(1+h)-(-1)}{h}$$

$$=\lim\limits_{h\to0^-}\dfrac{-5h}{h}=-5$$

$$\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}$$

$$=\lim\limits_{h\to0^+}\dfrac{3(1+h)-4(1+h)^2-(-1)}{h}$$

$$=\lim\limits_{h\to0^+}\dfrac{3+3h-4-8h-4h^2+1}{h}$$

$$=\lim\limits_{h\to0^+}(-5-4h)=-5$$

$$\lim\limits_{h\to0^-}\dfrac{f(1+h)-f(1)}{h}=-5$$

$$=\lim\limits_{h\to0^+}\dfrac{f(1+h)-f(1)}{h}$$

Then

$$\lim\limits_{h\to0}\dfrac{f(1+h)-f(1)}{h}=-5$$

The function $f(x)$ is derivable at $x=1,$ and

$$f'(1)=-5$$