Computing a function and determining if continuous or not

Question

Let $f$ be given by: $$ f(x)=\begin{cases} x^2 & x \leq 2\\ 2x & 2< x< 4\\ \sqrt{x} &x\geq 4 \end{cases} $$

A: Compute $f(1)$, $f(2)$, $f(3)$, $f(4)$, and $f(5)$.

B: At what points if any is $f(x)$ discontinuous?

Im struggling with where to calculate in the formula.

Thank You!

Answer 1

Just apply the definition, for example

• $f(1) = 1^2=1$
• $f(2) = 2^2=4$
• $f(3) = 2\cdot 3=6$
• $f(4) = \sqrt 4=2$

Note also that

• $\lim_{x\to 4^-} f(x)\neq \lim_{x\to 4^+} f(x)$

Answer 2

As for when $f$ is discontinuous, you just need to check the points $2$ and $4$. Since $x^2$, $2x$, and $\sqrt{x}$ are all continuous, you have that $f$ is continuous on $(-\infty,2)\cup(2,4)\cup(4,\infty)$.

At $x=2$, you have $$\lim_{x\rightarrow 2^-}f(x)=\lim_{x\rightarrow 2^-}x^2=4$$ and $$\lim_{x\rightarrow 2^+}f(x)=\lim_{x\rightarrow 2^+}2x=4$$ Since these limits agree, $f$ is continuous at $x=2$. Now, at $x=4$, you have $$\lim_{x\rightarrow 4^-}f(x)=\lim_{x\rightarrow 4^-}2x=8$$ and $$\lim_{x\rightarrow 4^+}f(x)=\lim_{x\rightarrow 4^+}\sqrt{4}=2$$ Since this limits do NOT agree, $f$ is discontinuous at $x=4$.