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March 23rd, 2018, 07:32 PM   #1
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Which of the following functions are differentiable at $x = 1$ ?

Which of the following functions are differentiable at $x = 1$ ?
a) $f(x) = ||x| - 1/2|$
b) $f(x) = max${$|x-1|, |x+1|$}
c) $f(x) =|x-1| + e^x$
d) $f(x) = |e^x -1| $

I have tried solving above functions but I didnt get any of them differentiable. I have doubt in solving option (a) incorrectly. Someone help on this!!!
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March 23rd, 2018, 07:59 PM   #2
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Answer is (d). Now can you see why ?
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March 23rd, 2018, 08:01 PM   #3
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When you graph (a) what does it do at x = 1?
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March 23rd, 2018, 08:04 PM   #4
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The absolute value function $|x|$ has a derivative everywhere except where it is equal to zero. So you should be looking to find out whether any of the four functions contain an absolute value function that has value zero at $x=1$.
  1. The inner absolute value function has derivative except at $x=0$, so it's smooth for $x \gt 0$, where $\big| |x| - \frac12 \big| = \big| x - \frac12 \big|$. Where does this function not have a derivative?
  2. $\max{(|x-1|, |x+1|)} = |x+1|$ for all $x \gt 0$. And $|x+1| = x+1$ for all $x \gt -1$.
  3. Treat each half of the function separately.
  4. When does $e^x-1 = 0$?

Last edited by v8archie; March 23rd, 2018 at 08:06 PM.
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March 23rd, 2018, 09:17 PM   #5
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Quote:
Originally Posted by SDK View Post
Answer is (d). Now can you see why ?
I assume $|e^x - 1| $is same as$ |x|$. So I took $-x$ (i.e.,$ -e^x+1$) when $x \leq 1 $ and $ x$(i.e., $e^x-1$) when $x \geq 1$. In this way they are neither differentiable nor continuous.
Let me know how to do it 😢
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March 24th, 2018, 06:00 AM   #6
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Quote:
Originally Posted by Lalitha183 View Post
I assume $|e^x - 1| $is same as$ |x|$. So I took $-x$ (i.e.,$ -e^x+1$) when $x \leq 1 $ and $ x$(i.e., $e^x-1$) when $x \geq 1$. In this way they are neither differentiable nor continuous.
Let me know how to do it 😢
If you mean, by "is the same as |x|", that you can treat it in the same way, yes.

For any positive x, $e^x> 1$ so $e^x- 1> 0$ and $|e^x- 1|= e^x- 1$ which is differentiable. This problem would be more complicated if the question were about x= 0 rather than x= 1.
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March 27th, 2018, 08:05 AM   #7
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Quote:
Originally Posted by Country Boy View Post
If you mean, by "is the same as |x|", that you can treat it in the same way, yes.

For any positive x, $e^x> 1$ so $e^x- 1> 0$ and $|e^x- 1|= e^x- 1$ which is differentiable. This problem would be more complicated if the question were about x= 0 rather than x= 1.
I have finally solved all 4 problems. And the results are like this :
1. L.H.L = R.H.L = $ \frac{1}{2} $
L.H.D = R.H.D = $1$ and $f(1) = \frac{1}{2} $

2. L.H.L = R.H.L = $2$
L.H.D = R.H.D = $0$ and $f(1) = 2 $

3. L.H.L = R.H.L = $e$
L.H.D = R.H.D = $0$ and $f(1) = e > 1$

4. L.H.L = R.H.L = $e - 1$
L.H.D = R.H.D = 0 and $f(1) = e-1$
In this, $e-1 >0$ then how will it be equal to $0$ to say $|e^x-1| $ is differentiable at $x = 1$
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March 29th, 2018, 07:38 AM   #8
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You say you have solved all four problems but the original problem was to determine whether or not each function is differentiable at x= 1 and nowhere do you answer that question!

a)f(x)= ||x|- 1/2|. For x larger than 1/2, |x|= x and x- 1/2> 0 so ||x|- 1/2|= x- 1/2 for all x> 1/2. Near x= 1, ||x|- 1/2|= x- 1/2 which is differentiable (and has derivative 1) at x= 1.

b)f(x)= max(|x- 1|, |x+1|). For x larger than 1, both x- 1 and x+ 1 are positive so |x- 1|= x- 1 and |x+ 1|= x+1. Obviously x+1 is larger than x- 1 so for x> 1, f(x)= x+ 1. If x< 1 then x- 1 is negative so |x- 1|= -(x- 1)= 1- x. So for x< 1, but close to 1, the function is max(1- x, x+ 1)= x+ 1 also. The "difference quotient" is (f(1+ h)- f(1))/h= h/h= 1 for x>1 or x< 1. The derivative at x= 1 is the limit of that as h goes to 0, 1. The function is differentiable at x= 1 (and has derivative 1).

c) f(x)= |x- 1|+ e^x. For x larger than 1, |x- 1|= x- 1, for x less than 1, |x- 1|= -(x- 1)= 1- x. The difference quotient, (f(1+h)- f(1))/h is, for h positive, so that 1+h> 1, (1+ h- 1+ e^{1+h}- e)/h= 1+ (e^{1+ h}- e)/h. As h goes to 0, that goes to 1+ 1= 2. For h negative, so that 1+h< 1, the difference quotient is (1- (1+h)+ e^{1+h}- e)/h= -1+ (e^{1+h}- e)/h. As h goes to 0, that goes to -1+ 1= 0. The two limits are not the same so this function is not differentiable at x= 1.

d) f(x)= |e^x- 1|. For x close to 1, e^x is close to e and e- 1> 0 so for x close to 1, f(x)= e^x- 1. That is differentiable so f(x) is differentiable for x= 1 (and the derivative is e- 1).
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