Show something is continuous but not differentiable? 10 points!?

Question

show that

f(x) = { x^2 - 5 , x <_ 1

. { x - 5 , x > 1

is continuous but not differentiable at x = 1.

Thanks in advance!!!

also, can you define what differentiable means?

Answer 1

The derivative of one side becomes 2x, as x approaches 1 from the right.  From the left, the derivative is x.

As x approaches 1, you then have a value of 1 from the left and 2 from the right.  Since the two limits are not equal, the function is not differentiable at x = 1.  There is a &quot;sharp point&quot; in the graph.

Answer 2

1) The derivative of the function that you have given is 1 for x &gt; 1 and 2 for x&lt;=1.  In order for a function to be differentiable at a point, the following must be true:

In order for a function f(x) to be differentiable at a point - call it x1 - the limit as delta x (the change in x) approaches 0 of:

f(x1 + delta x) - f(x1)

---------------------------------

delta x

must exist.  In this case the limit from the left and the limit from the right (I am talking about the graph now) are not equal.  Hence no unique such limit exists and your function is not differentiable.

2) In order for a function to be differentiable it must be differentiable at every point in its domain.  I am copying the above:

In order for a function f(x) to be differentiable at a point - call it x1 - the limit as delta x (the change in x) approaches 0 of:

f(x1 + delta x) - f(x1)

---------------------------------

delta x

must exist.

That must be true throughout the domain of f(x) for f(x) to be diffentiable.

Answer 3

finding left hand limit and right hand limit

LHL        lim    f(x)

              x-&gt;1-

             lim   x^2 -5      =       -4

             x&gt;1-  

RHL       lim    f(x)

              x-&gt;1+

             lim   x -5      =       -4

             x&gt;1+

F(1)  =  -4

NOW SINCE  LHL = RHL =F(1) SO FN IS CONTINUOUS AT X=1

NOW FIND LHD  AND RHD

lim   (f (x+h) - f (x)) / h           =     lim ( f (x+h) - f(x) ) /h

h-&gt;0-                                  h-&gt;0+

lim( (1+h)^2 - 5 - 1^2 + 5 ) / h           =     lim  (1+h - 5  - 1^2 + 5 )/h

h-&gt;0-                                             h-&gt;0+

on solving u get

  LHD  as  2

  RHD  as 1

since lhd is not equal to rhd so ur function is not differentiable at x=1.

well as far as i think differentiability at a point means that when u approach the pt from either of the two sides u must get the same value.