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January 24th, 2019, 03:48 AM  #1 
Member Joined: Feb 2018 From: Iran Posts: 30 Thanks: 3  Derivatives The answer in my book is 5 but if we put 5 instead of x all of them are differentiable 
January 24th, 2019, 04:02 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,298 Thanks: 1971 
$x^2  4$ isn't differentiable at Â±2 (two values of $x$), $\text{sgn}(x^3 + x)$ isn't differentiable at 0, etc.

January 24th, 2019, 04:10 AM  #3 
Senior Member Joined: May 2016 From: USA Posts: 1,306 Thanks: 549 
You are misreading the question. It is not asking for the actual values of x where f(x) is not differentiable. It is asking how many such values there are.

January 24th, 2019, 05:59 AM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 559 Thanks: 324 Math Focus: Dynamical systems, analytic function theory, numerics 
Unrelated comment: This is truly a terrible question. Whoever assigned it isn't paying attention. The only reasonable answer is 3, not 5. It only makes sense to talk about differentiability of a function on its domain. Functions don't "see" anything that isn't in their domain. Both of the following sentences make an equal amount of sense. 1. $f$ is not differentiable at $x = \frac{1}{\sqrt{2}}$. 2. $f$ is not differentiable at $x = $ 1998 Honda Accord. Last edited by skipjack; January 24th, 2019 at 12:52 PM. 
January 24th, 2019, 02:28 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,684 Thanks: 659 
$x^24=0$, 2 values. $x^3+x=0$, 1 real value. $2x^21=0$, 2 values. Total 5 (real) values. 
January 24th, 2019, 03:06 PM  #6 
Senior Member Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 
It is enough to know how to find the derivative , then the domain can be found easily .


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