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 March 21st, 2018, 06:00 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Range of Continuous & Differentiable Sets Let $f:R -> R$ and $f(x) = |x-1|+|x-2|$. Let $S1 =${$x| f$is continuous at $x$} and $S2 =${$x| f$ is differentiable at $x$} then a) $S1 = R , S2=R$ b) $S1 = R -${$1,2$}$, S2 = R$ c) $S1 = R , S2 = R -${$1,2$} d) $S1 = R -${$1,2$}$, S2 = R -${$1,2$} Which of the following is true ? I have seen individual questions like $f(x) = |x-1|$ where it is not continuous at $x=1$. Simillarly, $f(x) = |x-2|$ where it is not continuous at $x=2$. March 21st, 2018, 09:01 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,836 Thanks: 1479 for $x \ge 2$ ... $f(x) = (x-1)+(x-2) = 2x-3$ $f'(x) = 2$ for $1 < x < 2$ ... $f(x) = (x-1) - (x-2) = 1$ $f'(x) = 0$ for $x \le 1$ ... $f(x) = -(x-1)-(x-2) = -2x+3$ $f'(x) = -2$ $f(x)$ is continuous over $\mathbb{R}$ $f(x)$ is differentiable over $\mathbb{R} - \{1,2\}$ Thanks from Lalitha183 March 22nd, 2018, 04:13 AM #3 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2 Thank you so much. We have to redefine the function to check continuity & differentiability since the limits have not given. And after checking continuity & differentiability over 1 & 2, it has given the clear answer. March 24th, 2018, 06:14 AM   #4
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 Originally Posted by Lalitha183 Let $f:R -> R$ and $f(x) = |x-1|+|x-2|$. Let $S1 =${$x| f$is continuous at $x$} and $S2 =${$x| f$ is differentiable at $x$} then a) $S1 = R , S2=R$ b) $S1 = R -${$1,2$}$, S2 = R$ c) $S1 = R , S2 = R -${$1,2$} d) $S1 = R -${$1,2$}$, S2 = R -${$1,2$} Which of the following is true ? I have seen individual questions like $f(x) = |x-1|$ where it is not continuous at $x=1$. Simillarly, $f(x) = |x-2|$ where it is not continuous at $x=2$.
No, $|x- 1|$ is continuous at x= 1. It is not differentiable there.

If x< 1 then both x- 1 and x- 2 are negative so f(x)= -(x- 1)- (x- 2)= -2x+ 3.

If 1< x< 2 then x- 1 is positive and x- 2 is still negative so f(x)= x- 1- (x- 2)= 1.

If x> 2 then both x- 1 and x- 2 are positive so f(x)= x- 1+ x- 2= 2x- 3.

The limit, as x goes to 1 "from below", is -2(1)+ 3= 1 and "from above" is 1 so f is continuous there. The derivative of f, for x< 1,is just the slope of -2x+ 3, -2. For 1< x< 2, f is constant so the derivative is 0. Those are not the same so f is not differentiable at x= 1.

The limit,as x goes to 2 "from below", is 1 and "from above" is 2(2)- 3= 1 so the function is continuous at x= 2. The derivative of x, for 1< x< 2, is 0. The derivative of f, for x> 2, is 2. Those are not the same so f is not differentiable at x= 2.

(Note: the derivative of a continuous function is NOT necessarily continuous itself but must satisfy the "intermediate value property" so that the two right and left sided slopes not being the same is sufficient to say that the derivative does not exist.) Tags continuous, differentiable, range, sets Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Lotta Real Analysis 0 October 8th, 2017 05:09 AM Erdnuss Calculus 4 February 2nd, 2015 03:18 PM wannabe1 Real Analysis 5 November 7th, 2010 03:14 PM chloe561 Real Analysis 2 March 20th, 2009 06:04 AM clooneyisagenius Calculus 2 February 29th, 2008 06:29 AM

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