in Calculus recategorized by
515 views
0 votes
0 votes

Consider the function $$f(x) = \begin{cases} \int_0^x \{5+ \mid 1-y \mid \} dy & \text{ if } x>2 \\ 5x+2 & \text{ if } x \leq 2 \end{cases}$$ Then

  1. $f$ is not continuous at $x=2$
  2. $f$ is continuous and differentiable everywhere
  3. $f$ is continuous everywhere but not differentiable at $x=1$
  4. $f$ is continuous everywhere but not differentiable at $x=2$
in Calculus recategorized by
by
515 views

1 Answer

2 votes
2 votes
$$f(x)=\begin{cases}
 \displaystyle{\int_{0}^{x}5+|1-y| \ \ \mathrm{d}y} & \text { if } x>2 \\[2ex]
5 x+{2} & \text { if } x \leq 2
\end{cases}$$

$$f'(x)=\begin{cases}
 5+|1-x| & \text { if } x>2 \\[2ex]
5  & \text { if } x \leq 2
\end{cases}$$
Using Newton-Leibnitz rule for differentiation
  

for checking the  differentiability: $$f'(2^+)=\lim_\limits{h \to 0} 5+|1-(2+h)|=6 $$
And
$$f'(2^-)=5$$
Hence,

the function is not differentiable ($\because f'(2^+)\neq f'(2^-) $) .
edited by

2 Comments

The first task should have been to check whether the function is continuous. Indeed, it is continuous. Hence, is the correct answer.

0
0
It is not continuous, after solving for x>2 condition we get  $((x^2)/2)+4x+1$  then f($2^+$) =11,

f($2^-$) =$5x+2$=12,

f($2^+$)$≠$f($2^-$)

So it is not continuous.

It would have been continuous if the f($2^-$)=$5x+1$, but it is not
0
0

Related questions