I am doing my calculus homework and two problems are holding me up.(adsbygoogle = window.adsbygoogle || []).push({});

The first says:

Using one-sided derivatives, show that the function f(x) =

x^3, x_<_1

3x, x>1

does not have a derivative at x=1

Now it is painfully obvious that the function is not continuous at x=1. however, i am not entirely sure that is the answer that the book wants. the slopes from both sides seem to be 3, so how can it not have a derivative (aside from the continuity issue)? A similar (but with continuity) problem follows:

Secondly (i will try to answer the ones i have gotten so far):

let f(x) =

x^2, x_<_1

2x, x>1

a) find f'(x) for x<1........................i think this is 2x

b) find f'(x) for x>1........................2

c) find lim (x-->1-) f'(x).................2

d) find lim (x-->1+) f'(x).................2

e) does lim (x-->1) f'(x) exist? explain

f) use the def to find the left-hand derivative of f at x=1 if it exists....same as (c)=2

g) use the def...right-hand deriv.....same as (d)=2

h) does f'(1) exist? explain

according to the rules of derivatives, if the left and right-hand derivatives are the same at a point, then that point has a derivative (assuming continuous). however, it seems to me that, at x=1, there would be a bit of a "jagged edge," somewhat like an absolute value point. therefore, how could a derivative be found?

thanks in advance. feel free to tell me i am horribly wrong in all aspects of my answer.

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# Derivative of Piece-Wise Function

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