search
Log In
2 votes
141 views

Let the function $f(x)$ be defined as $f(x)=\mid x-1 \mid + \mid x-2 \:\mid$. Then which of the following statements is true?

  1. $f(x)$ is differentiable at $x=1$
  2. $f(x)$ is differentiable at $x=2$
  3. $f(x)$ is differentiable at $x=1$ but not at $x=2$
  4. none of the above
in Calculus
retagged by
141 views

1 Answer

0 votes

Option D

The f(x) is neither differentiable at x=1 nor at x=2 because f(x) is pointed at both x=1 and x=2, though it is continuous in (-inf,+inf)

Related questions

0 votes
1 answer
1
149 views
If $f(x) = \sin \bigg( \dfrac{1}{x^2+1} \bigg),$ then $f(x)$ is continuous at $x=0$, but not differentiable at $x=0$ $f(x)$ is differentiable at $x=0$, and $f’(0) \neq 0$ $f(x)$ is differentiable at $x=0$, and $f’(0) = 0$ None of the above
asked Sep 23, 2019 in Calculus Arjun 149 views
0 votes
1 answer
2
113 views
Let $f(x) = \dfrac{x}{(x-1)(2x+3)}$, where $x>1$. Then the $4^{th}$ derivative of $f, \: f^{(4)} (x)$ is equal to $- \frac{24}{5} \bigg[ \frac{1}{(x-1)^5} – \frac{48}{(2x+3)^5} \bigg]$ $\frac{24}{5} \bigg[ – \frac{1}{(x-1)^5} + \frac{48}{(2x-3)^5} \bigg]$ $\frac{24}{5} \bigg[ \frac{1}{(x-1)^5} + \frac{48}{(2x+3)^5} \bigg]$ $\frac{64}{5} \bigg[ \frac{1}{(x-1)^5} + \frac{48}{(2x+3)^5} \bigg]$
asked Sep 23, 2019 in Others Arjun 113 views
2 votes
1 answer
3
243 views
Let $a_n=\bigg( 1 – \frac{1}{\sqrt{2}} \bigg) \cdots \bigg( 1 – \frac{1}{\sqrt{n+1}} \bigg), \: n \geq 1$. Then $\underset{n \to \infty}{\lim} a_n$ equals $1$ does not exist equals $\frac{1}{\sqrt{\pi}}$ equals $0$
asked Sep 23, 2019 in Calculus Arjun 243 views
4 votes
4 answers
4
346 views
$\underset{x \to \infty}{\lim} \left( \frac{3x-1}{3x+1} \right) ^{4x}$ equals $1$ $0$ $e^{-8/3}$ $e^{4/9}$
asked Sep 23, 2019 in Calculus Arjun 346 views
...