Find values for this piecewise function so it can be differentiable

[zaddyzad]

Homework Statement

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.

The Attempt at a Solution

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

 

[SammyS]

Homework Statement

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

a) Find all values of a, b, c and d that make the function f differentiable on the domain -∞<x<∞

b) Using the values found 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.

The Attempt at a Solution

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

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.

 

[zaddyzad]

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

 

[zaddyzad]

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.

f'(x) = 2ax + b
f'(x) = (-2/x) cos(1/x)... I guess f'(0) doesn't exist for this?

 

[zaddyzad]

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

So B = 0...

 

[zaddyzad]

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 isn't a answer at that point ?

 

[haruspex]

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

That's not how the product rule works.

 

[SammyS]

f'(x) = 2ax + b , for x < 0
f'(x) = 2x cos(1/x)(-1/x^2) , for x > 0

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). )