Expert: Scotto - 7/21/2009

Question
QUESTION: COnsider the function f:R->R defined by:

f(x)= ((1+x)/2) if x<1,
         1     if x=1,
      sqrt(x)  if x>1

a) Find f'(1)
b) Find f"(1)

c)I would also like to know the function for f'(x) and for f"(x)

Thank you

ANSWER: The function f(x) = 1/2 for x < 1;
f(x) = 1 at x = 1; and √x is x > 1.

x<1 : f'(x) = 0 and f"(x) = 0.
x>1 : f'(x) = 1/(2√x) and f"(x) = -1/(4√x^3).

For x<1, f'(x) = 0 and for x>1, f'(x) = 1/(2*1) = 1/2.
From these two ways of finding f'(1),
it can be seen to have different values from eith side.

For x>1, f"(x) = 0 and for x>1 f"(x) = -1/(4*1) = -1/4.
Therefore it can also be said that f"(x) does not exist at x=1.

Note that the values of f'(x) and f"(x) were given above for x<1 and
x>1, but at x=1, there is no derivative.


---------- FOLLOW-UP ----------

QUESTION: f(x)=((1+x)/2)=(.5+.5x) [One half, times, One Plus x]

Not

f(x)=(1/2)

Thank you

Answer
Lets just ignore the fact I sent that document to you.
I think I'm getting so I don't even reread them all now and then.
Leads to mistake, you know?
Of course you know - you got one.

For x < 1, f'(x) = 1/2 and f"(x) = 0.
For x > 1, f'(x) = 1/(2√x) and f"(x) = -1/(4√x^3).

As x goes up to 1, f'(1) would be 1/2, and f"(1) would be 0.
As x goes down to 1, f'(1) would be 1/2 and f"(1) would be -1/4.

From here, it can be seen that f'(x) = 1/2 as x approaches 1,
since it does that as the limit from both sides.

However, for f"(x), from the left the limit is 0 and from the right the limit would be -1/4.  Therefore f"(1) does not exist.

Now that looks like it's correct.