Absolute value and piecewise functions

[939]

Homework Statement

Find if continuous and differentiable. I am having problems with the differentiable part.

Homework Equations

f(x) = x² + 3, |x| ≤ 1
f(x) = |x| + 3, |x| > 1

The Attempt at a Solution

(1)^2 + 3 = 4
|1| + 3 = 4
∴ It is continuous

Now, rewriting it, you have...
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1 or x > 1
x + 3, x < -1 or x > 1

Finding the limits from left and right of 1...
From the left = 2x, = 2
From the right: can be both -1 and 1...

Where did I go wrong?

 

[Dick]

Homework Statement

Find if continuous and differentiable. I am having problems with the differentiable part.

Homework Equations

f(x) = x² + 3, |x| ≤ 1
f(x) = |x| + 3, |x| > 1

The Attempt at a Solution

(1)^2 + 3 = 4
|1| + 3 = 4
∴ It is continuous

Now, rewriting it, you have...
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1 or x > 1
x + 3, x < -1 or x > 1

Finding the limits from left and right of 1...
From the left = 2x, = 2
From the right: can be both -1 and 1...

Where did I go wrong?

No, you've got
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1
x + 3, x > 1

Worry about the two point x=1 and x=(-1) separately.

 

[939]

No, you've got
x² + 3, x ≥ -1 or x ≤ 1
-x + 3, x < -1
x + 3, x > 1

Worry about the two point x=1 and x=(-1) separately.

Thanks a lot!

Final question can |(2/3)(x)| be turned into (2/3)(-x) and (2/3)(x)?

 

[Dick]

Thanks a lot!

Final question can |(2/3)(x)| be turned into (2/3)(-x) and (2/3)(x)?

Sure. |2x/3| is -2x/3 if x<0 and 2x/3 if x>=0.