# Find values for this piecewise function so it can be differentiable

1. Mar 13, 2013

1. The problem statement, all variables and given/known data

Fx= ax^2 + bx + c -infiniti<x<0
= D x=0
=x^2 sin(1/x) - 2 0<x<infiniti

a) FInd all vlaues of a, b, c and d that make the function f differntiable on the domain -∞<x<∞

b) Using the values founbd in part a, determine lim x-> 0- f'(x), lim x->0+ f'(x) and f'(0). Is f'(x) continuous at x=0. Explain.

3. The attempt at a solution

I need help with part b).... a=all real numbers b= all real numbers c= -2 d = -2

2. Mar 13, 2013

### SammyS

Staff Emeritus
You need help with part a) also.
To be differentiable, the function must also be continuous. You've accomplished that with your choices of c and d .

To be differentiable at x=0, f'(0) must exist.

What is $\displaystyle \ \frac{d}{dx}(ax^2 + bx + c)\ ?$

What is $\displaystyle \ \frac{d}{dx}(x^2 \sin(1/x) - 2)\ ?$

b is not arbitrary. It must be some specific value.

3. Mar 13, 2013

f'(x) = 2ax + b
f'(x) = 2x cos(1/x)(-1/x^2)

4. Mar 13, 2013

f'(x) = 2ax + b
f'(x) = (-2/x) cos(1/x).... I guess f'(0) doesnt exist for this?

5. Mar 14, 2013

lim = 2a(0) + b = f'(0) = lim (-2/x) cos(1/x)
x-> 0- ..................................................... x->0+

So B = 0....

6. Mar 14, 2013

How do you find the
lim for the f'(x) of the sin function.
x-> 0+

Or do you say the derivative is not constant at x=0 b/c there isnt a answer at that point ?

7. Mar 14, 2013

### haruspex

That's not how the product rule works.

8. Mar 14, 2013

### SammyS

Staff Emeritus
Your f'(x) is wrong for x > 0.

You need to use the product rule. ( In other words, your derivative is incorrect for x2sin(1/x). )