Derivatives

Hi, I just started Calculus last week, and this is a question that appeared on tonight's homework.

The graph of f(x)=|x| has a corner at x=0. Discuss why f'(0) is undefined?

So im guessing find the derivative, Which i think is

f'(x)=\displaystyle\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}

f'(x)=\displaystyle\lim_{h\to 0} \frac{|x+h|-|x|}{h}

f'(x)=\displaystyle\lim_{h\to 0} \frac{|h|}{h}

But as h--> 0, that would create the whole fraction as undefined....

Am i doing this all wrong?

Reply 1

Necro Defain

11

As you approach from the right (h>0), f'(0)=1, as you approach from the left (h<0), f'(0)=-1, the two limits are not equal so you cannot say "the limit of the difference quotient at 0 is ...), so the derivative is undefined

Reply 2

nk9230

OP

Necro Defain

As you approach from the right (h>0), f'(0)=1, as you approach from the left (h<0), f'(0)=-1, the two limits are not equal so you cannot say "the limit of the difference quotient at 0 is ...), so the derivative is undefined

basically because the limit does not exist, the derivative does not exist?

Reply 3

Swayum

18

nk9230

basically because the limit does not exist, the derivative does not exist?

Both the left and right hand limits must be equivalent for the derivative to be defined

Reply 4

nuodai

17

nk9230

basically because the limit does not exist, the derivative does not exist?

The derivative is a function that describes a slope at any one point of a graph, so it would make sense that any one point only has one gradient (that is,

Unparseable latex formula:

\mbox{f}'(a) = \mbox{f}'(b)

if

a = b

), but if you have two gradients at one point, this is essentially saying that

-1 = 1

, which is a contradiction.

Derivatives

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