The number of points at which f(x)=|x^2-4| is not differentiable.
Answers
Answered by
1
f(2+t) = | t² + 4t + 4 - 4 |
= | t (t+4) |
f(2) = 0
f(2+t) - f(2) = |t² - 4t|
-------------- ----------
t t
limit as t -> 0+ and 0- have different values
So at t = 0 and x=2, derivative does not exist
= | t (t+4) |
f(2) = 0
f(2+t) - f(2) = |t² - 4t|
-------------- ----------
t t
limit as t -> 0+ and 0- have different values
So at t = 0 and x=2, derivative does not exist
mathukkutty:
Thank you but Why did we take (t+2=x)?
to be able to prove by way of : definition of derivative in terms of limits
Thank you
Similar questions