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March 23rd, 2018, 07:32 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4  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) $f(x) = max${$x1, x+1$} c) $f(x) =x1 + e^x$ 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!!! Thanks 
March 23rd, 2018, 07:59 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 669 Thanks: 440 Math Focus: Dynamical systems, analytic function theory, numerics 
Answer is (d). Now can you see why ?

March 23rd, 2018, 08:01 PM  #3 
Senior Member Joined: Aug 2012 Posts: 2,409 Thanks: 753 
When you graph (a) what does it do at x = 1?

March 23rd, 2018, 08:04 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
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 $x=1$.
Last edited by v8archie; March 23rd, 2018 at 08:06 PM. 
March 23rd, 2018, 09:17 PM  #5 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4  
March 24th, 2018, 06:00 AM  #6  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
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.  
March 27th, 2018, 08:05 AM  #7  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 242 Thanks: 4  Quote:
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 $f(1) = e1$ In this, $e1 >0$ then how will it be equal to $0$ to say $e^x1 $ is differentiable at $x = 1$  
March 29th, 2018, 07:38 AM  #8 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
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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