Function: continuous?

[JH_33]

I was given the piece-wise function:

f(x) =

{ x^2 + 2, x < 1
{ 2x+1 , x ≥ 1

The question asks if the function is continuous
and/or differentiable at x=1

I would say that the function is not continuous but it is differentiable.
Is my thinking correct on this problem?

Thanks for the help.

 

[pka]

Given \(\displaystyle f(x) = \left\{ {\begin{array}{c c}
{x^ 2 + 2,} & {x < 1} \\
{2x + 1} & {1 \le x} \\
\end{array}} \right.\)

If it true that \(\displaystyle \ {{\rm{lim}}}\limits_{x \to 2^ + } f(x) = \ {{\rm{lim}}}\limits_{x \to 2^ - } f(x) = f(1)\) then it is continuous!

You have the other backwards.
If a function is differentiable at a point it must be continuous there.

 

[stapel]

I would say that the function is not continuous but it is differentiable.
Is my thinking correct on this problem?

Since being differentiable at a point means that the function is also continuous at that point, it would be difficult for your solution to be correct.

On what basis do you think the function is not continuous at x = 1? You've drawn the graph...? You've evaluated each half's rule at x = 1...?

On what basis do you think the function has a well-defined derivative at x = 1? You've compared the slopes of the two pieces on your graph...? You've evaluated the derivative formulas for each half at x = 1...?

Thank you.

Eliz.

 

[JH_33]

Ok... thanks for the help.

I guess im not thinking to clearly tonight!