MHB Is this function differentiable at c=1?

[cbarker1]

f'(x) is equal to the limit of x approaches to c of the (f(x)-f(c))/(x-c), and c=1.

f(x)=\begin{array}{cc}x,&\mbox{ if }
x\leq 1\\x^2, & \mbox{ if } x>1\end{array}The function is a piece wise function.Thanks

CBarker1

 

[MarkFL]

You need to investigate two one-sided limits. Can you state these two limits?

 

[Fermat1]

f'(x) is equal to the limit of x approaches to c of the (f(x)-f(c))/(x-c), and c=1.

f(x)=\begin{array}{cc}x,&\mbox{ if }
x\leq 1\\x^2, & \mbox{ if } x>1\end{array}The function is a piece wise function.Thanks

CBarker1

The function is not differentiable at 1 because the limit from the left is 1 but the limit from the right is 2.

 

[MarkFL]

The function is not differentiable at 1 because the limit from the left is 1 but the limit from the right is 2.

(Wait) I was actually hoping the OP could state the two one-sided limits and then discover these values on their own. (Nod)

 

[cbarker1]

I have worked out the problem. I see the limits are not equal to each other. Thank you.

 

[Prove It]

For future reference:

A function is continuous at a point if the function is defined at that point, and also the left hand and right hand limits are both equal to that function value.

A function is differentiable at a point if the function is continuous at that point, and also if the derivative is continuous at that point.