Question

17 mark for review
f(x)=\begin{cases}2^{x}&\text{for }0 < x < 1\\frac{1}{2}x^{2}-x+\frac{5}{2}&\text{for }1 < x < 2end{cases}
let (f) be the function defined above. which of the following statements is true?
a (f) is continuous at (x = 1).
b (f) is not continuous at (x = 1) because (f(1)) does not exist.
c (f) is not continuous at (x = 1) because (lim_{x
ightarrow1^{-}}f(x)
eqlim_{x
ightarrow1^{+}}f(x)).

Explanation:

Step1: Calculate left - hand limit

We find $\lim_{x
ightarrow1^{-}}f(x)$. Since for $0 < x<1$, $f(x)=2^{x}$, then $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}2^{x}=2^{1} = 2$.

Step2: Calculate right - hand limit

We find $\lim_{x
ightarrow1^{+}}f(x)$. Since for $1 < x<2$, $f(x)=\frac{1}{2}x^{2}-x+\frac{5}{2}$, then $\lim_{x
ightarrow1^{+}}f(x)=\frac{1}{2}(1)^{2}-1+\frac{5}{2}=\frac{1 - 2+5}{2}=2$.

Step3: Check function value

The function $f(x)$ is not defined at $x = 1$ because the domain of the first - part is $0ightarrow a}f(x)$ must exist and $\lim_{x
ightarrow a}f(x)=f(a)$. Here, although $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x) = 2$, $f(1)$ does not exist.

Answer:

B. $f$ is not continuous at $x = 1$ because $f(1)$ does not exist.