Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{1}{5}(2x^2+3)& -\infty <x\le 1\\ 6-5x& 1<x<3\\ x-3& 3\le x<\infty \end{array}$

• A. $f$ is continuous at $x=1$
• B. $f$ is discontinuous at $x=1$
• C. $f$ is continuous at $x=3$
• D. $f$ is discontinuous at $x=2$

Answer 1

The correct option is A $f$ is continuous at $x=1$

Checking for continuity at $x=1$:
${lim}_{x\to 1^{-}}f\left(x\right)=f\left(1\right)=\frac{1}{5}(2\cdot 1^2+3)=1{lim}_{x\to 1^{+}}f\left(x\right)=6-5\cdot 1=1$
Since ${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1^{+}}f\left(x\right)=f\left(1\right)$, the function is continuous at $x=1$.
Checking for continuity at $x=3$:
${lim}_{x\to 3^{-}}f\left(x\right)=6-5\cdot 3=-9{lim}_{x\to 3^{+}}f\left(x\right)=f\left(3\right)=3-3=0$
Since ${lim}_{x\to 3^{-}}f\left(x\right)\ne {lim}_{x\to 3^{+}}f\left(x\right)$, the function is not continuous at $x=3$.
$f\left(x\right)$ is an algebraic function that does not change its definition at $x=2$, so it is continuous at $x=2$.
The correct answer is option A.