Solving f(x): Finding a & b for Continuity/Differentiability

[Fusilli_Jerry89]

Let f be the function defined as:

f(x) = { x+2 , x < 2
{ ax^2+bx, x >(or equal to) 2

a) what is the relationship between a and b?
I got 2a+b=2
b) find the unique values of a and b that will make f both continuous and differentiable.

i substitued (2-2a) into b and got down to y=ax^2+2x-2ax, but now what?

 

[berkeman]

I moved this thread to the homework forums. Jerry -- you've used the homework forums before, so I know that you know the rules. This is the 2nd thread of yours that I've had to move tonight. So even though you showed some work in this post (none in the previous post), I'm going to issue some warning points. Please keep homework posts in the HW forums.

 

[berkeman]

Well, except with the new PF forum changes, I don't see the usual WARN button. Okay, you skated this time -- next time I look harder for the new WARN button placement...

 

[Pyrrhus]

Well so far, you considered the continuity condition, but what about the differentiability condition?

 

[HallsofIvy]

Let f be the function defined as:

f(x) = { x+2 , x < 2
{ ax^2+bx, x >(or equal to) 2

a) what is the relationship between a and b?
I got 2a+b=2
b) find the unique values of a and b that will make f both continuous and differentiable.

i substitued (2-2a) into b and got down to y=ax^2+2x-2ax, but now what?

Are you sure you have quoted the problem correctly?
If
"Let f be the function defined as:

f(x) = { x+2 , x < 2
{ ax^2+bx, x >(or equal to) 2"
there doesn't have to be ANY particular relationship between a and b!
HOW did you get 2a+ b= 2?

IF f is continuous at x= 2, then f(2)= 2+ 2= 4= 4a+ 2b so you get 2a+ b= 2 but that isn't given until part b.

What is the derivative of x+ 2 at x= 2? What is the derivative of ax²+ bx at x= 2?


(Note: If a function is differentiable at x= 2, it is not necessarily the case that the derivative is continuous at x= 2. However, any derivative must satisfy the a "intermediate value property" so if the two "one-sided" limits exist, they must be the same. THAT means that the two values above must be the same.)

By the way, the word is "PIECEwise".

 

[physics.guru]

Also keep in mind, that a continuous function is not necessarily differentiable at a point x. Case in point: f(x) = | x | at x = 0

Here we see that f(x) satisfies the 3 conditions necessary for continuity:

1. f(0) is defined

2. $$\lim_{x\rightarrow 0} | x |$$ exists

3. $$\lim_{x\rightarrow 0} = f(0)$$

Yet, f(x) is not differentiable at x = 0 (differentiable everywhere else).

If f is differentiable at a, then f is continuous at a. Unfortunately, this does not mean if f is continuous at a, f is differentiable at a. As seen above.