Question

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

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

Answer 1

Answer: (A), (D)

discontinuous at $x=3$
continuous at $x=1$

Checking for continuity at $x=1$:
$f\left(1\right)=\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{-}}{lim}\frac{1}{5}(2x^2+3)=\frac{1}{5}(2\cdot 1^2+3)=1$
$\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(6-5x)=6-5\cdot 1=1$
The function is continuous at this point.
Checking for continuity at $x=3$,
$\underset{x\to 3^{-}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}(6-5x)=6-5\cdot 3=-9$
$f\left(3\right)=\underset{x\to 3^{+}}{lim}f\left(x\right)=\underset{x\to 3^{-}}{lim}(x-3)=3-3=0$
Since the two limits are not equal, the function is discontinuous at this point.