Answer to Question #283627 in Calculus for Rekha

Question

let

f(x) =1 + 2𝑥, 𝑥 ≤ 0

=3𝑥 − 2,0 < 𝑥 ≤ 1

= 2𝑥 ଶ − 1, 𝑥 > 1

i) Check whether f is discontinuous. If yes, find where? ii) Give a rough sketch of the graph of f. 

Answer 1

$$f(x) = \begin{cases} 1+2x, x\leq 0 \\ 3x-2, 0<x\leq 1 \\ 2x-1, x> 1 \\ \end{cases}$$

i)

$$\lim\limits_{x\to 0^-}f(x)=\lim\limits_{x\to 0^-}(1+2x)=1+2(0)=1$$ $$\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^+}(3x-2)=3(0)-2=-2$$ $$\lim\limits_{x\to 0^-}f(x)=1\not=-2=\lim\limits_{x\to 0^+}f(x)$$ $$\lim\limits_{x\to 0}f(x) \text{does not exist}$$

The function $f(x)$ has an jump discontinuity at $x=0.$

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

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

The function $f(x)$ is discontinuous at $x=0.$

The function $f(x)$ has an jump discontinuity at $x=0.$

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

ii)