Conditions of differentiability (1 Viewer)

[lsdpoon1337]

Hey,

I'm having trouble with the following:

Find the value of f '(a) in each case where it exists.

f(x) = x, x>3 a=3 and a= -3
3, -3 <= x <= 3
x + 1, x < -3


My answer is:

f '(3) cannot be found since the left-hand derivative f(x)= 3 is 0 and the right-hand derivative, f(x)= x is 1 even though the function is continuous at x=a=3.

f '(-3) cannot be found since the left-hand derivative, f(x)= x + 1 does not exist at x= -3, even though f(-3) is defined as 3 using f(x)= 3

However, the answer in the 2 unit fitzpatick (question 6, exercise 14a) is:

0 when a=3

How is this so? Or am I right?

 

[roadrage75]

what's the question? are u trying to graph f(x)?

 

[undalay]

if the general limit does not exist, the function shouldn't be continuous ?

 

[roadrage75]

you seem right to me, i'm guessing the book is wrong.-- and, i guessing you've already done this, but if you draw the graph, it becomes very clear that it is not differentiable at a= 3