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 March 23rd, 2018, 07:19 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Differentiable or not Let $f(x) = x$ ; $x \in Q$ $0$; $x \notin Q$ The function at point $x = 0$ a) Is differentiable b) has the left derivative but not the right derivative c) has the right derivative but not the left derivative d) has neither the left derivative nor the right derivative Since both the functions are equal at $x = 0$, I think it is differentiable. Please help! Last edited by skipjack; March 26th, 2018 at 01:02 AM.
 March 23rd, 2018, 07:57 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 578 Thanks: 345 Math Focus: Dynamical systems, analytic function theory, numerics What is the definition of differentiability?
March 23rd, 2018, 08:51 PM   #3
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 Originally Posted by SDK What is the definition of differentiability?
If the left hand derivative is equal to the right hand derivative equals to $f(a)$ then it is differentiable.

March 23rd, 2018, 09:07 PM   #4
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 Originally Posted by Lalitha183 If the left hand derivative is equal to the right hand derivative equals to $f(a)$ then it is differentiable.
If $f(x) = 1$ then the left and right hand derivatives at $0$ are $0$, but $f(0)$ = 1, right?

March 23rd, 2018, 09:45 PM   #5
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 Originally Posted by Maschke If $f(x) = 1$ then the left and right hand derivatives at $0$ are $0$, but $f(0)$ = 1, right?
I don't understand what does it mean by $f(x) = 1$ ? And how will $f(0) = 1$ ?

March 23rd, 2018, 11:48 PM   #6
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 Originally Posted by Lalitha183 I don't understand what does it mean by $f(x) = 1$ ? And how will $f(0) = 1$ ?
$f$ is the constant function that inputs any real number and always returns the value $1$. What is its derivative? What's its value at $x = 0$?

Last edited by Maschke; March 23rd, 2018 at 11:57 PM.

March 24th, 2018, 04:47 AM   #7
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 Originally Posted by Maschke $f$ is the constant function that inputs any real number and always returns the value $1$. What is its derivative? What's its value at $x = 0$?
The Left Hand Derivative = Right Hand Derivative = $0$. But $f(0) = 1$. So the function is not differentiable at $x=0$.

 March 24th, 2018, 05:54 AM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 The derivative of a function at a point is completely independent of the value of the function at that point, so your "but f(0) = 1" is irrelevant. The function f(x) = 1 has derivative 0 at any x. But the function in the original post has value x for any rational x, 0 for any irrational x. The difference quotient $\displaystyle \frac{f(0+ h)- f(0)}{h}$ is $\displaystyle \frac{0+ h- 0}{h}= 1$ for h rational and $\displaystyle \frac{0- 0}{h}= 0$ for h irrational. The limit, as h goes to 0, does not exist, so this function is not differentiable at x = 0. Last edited by skipjack; March 26th, 2018 at 01:05 AM.
March 24th, 2018, 08:58 AM   #9
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 Originally Posted by Lalitha183 The Left Hand Derivative = Right Hand Derivative = $0$. But $f(0) = 1$. So the function is not differentiable at $x=0$.
Can you graph $f(x) = 1$? You are confusing the definitions of continuity and differentiability. This example is so elementary that given your educational aspiration, you cannot afford to misunderstand it.

March 26th, 2018, 12:31 AM   #10
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 Originally Posted by Country Boy The derivative of a function at a point is completely independent of the value of the function at that point, so your "but f(0) = 1" is irrelevant. The function f(x) = 1 has derivative 0 at any x. But the function in the original post has value x for any rational x, 0 for any irrational x. The difference quotient $\displaystyle \frac{f(0+ h)- f(0)}{h}$ is $\displaystyle \frac{0+ h- 0}{h}= 1$ for h rational and $\displaystyle \frac{0- 0}{h}= 0$ for h irrational. The limit, as h goes to 0, does not exist, so this function is not differentiable at x = 0.
Which one do we take as RHD & LHD? Because to my question, answers are like is differentiable/ has L.H.D but not R.H.D/ has R.H.D but not L.H.D/ Neither L.H.D nor R.H.D.

Last edited by skipjack; March 26th, 2018 at 01:05 AM.

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